In this article, we shall study the vibration of vertical spring, when a mass is attached to it.

Oscillation of Mass Due to a Vertical Spring:

Let us consider light and elastic spring of length L suspended vertically from a rigid support. In a relaxed state the spring is unstretched. Let a small mass m be attached to its free end. Due to the action of deforming force mg, the spring extends. Let the extension in the spring be l.

By Hooke’s law, Restoring force ∝ extension

F ∝ – l

F = – k l   ……… (1)

Where k = constant called the force constant of the spring.

The negative sign indicates that the direction of restoring force (upward) is opposite to the direction of extension (downward). Let the mass be pulled down by small force ΔF and let the corresponding increase in length be y.

F + ΔF = – k (l   +  y)   ……… (2)

Subtracting equation (1) from (2)

(F + ΔF) – F = – k (l   +  y) + k l   .

ΔF = – k y

This is the restoring force which will create an acceleration in the mass when released.

Thus the acceleration in the mass is directly proportional to the displacement of the mass and opposite to it. But this is the defining character of the linear S.H.M. Thus when the pulled down mass is released it starts performing linear S.H.M. The time period of a particle performing linear S.H.M. in terms of force constant is given by

By Hooke’s Law F = – k l 

Considering magnitude only F = k l 

This restoring force is equal to the weight of the mass attached

mg = kl

Numerical Problems on Vibration of Vertical Spring:

Example – 01:

A load of 200 g increases the length of a light spring by 10 cm. Find the period of its vertical oscillations when a mass of one kg is attached to the free end of the spring. Take g = 10 m/s².

Given: Stretching load = F = 200 g = 200 x 10^-3 kg= 200 x 10^-3 x 10 = 2 N, Increase in length = l = 10 cm = 10 x 10^-2 m, mass attached = m = 1 kg, g = 10 m/s².

To Find: Period = T = ?

Solution:

Force constant k = F/l = 2 / 10 x 10^-2 = 20 N/m

Ans: Period of oscillation = 1.41 s

Example – 02:

A vertical light spring is stretched by 2 cm when a weight of 10 g attached to its free end. The weight is further pulled down by 1 cm and released. Compute its frequency and amplitude.

Given: Stretching load = F = 10 g = 10 x 10^-3 kg= 10 x 10^-3 x 9.8 = 9.8 x 10^-2 N, increase in length = l  = 2 cm = 2 x 10^-2 m, mass attached = m = 10 g = 10 x 10^-3 kg, g = 9.8 m/s², distance through which the mass is pulled down = y =  1 cm = 1 x 10^-2 m

To Find: Frequency = ?, amplitude = ?

Solution:

Force constant k = (9.8 x 10^-2)/( 2 x 10^-2) = 4.9 N/m

The amplitude = distance through which mass is pulled down.

amplitude = a = 1 cm = 0.01 m

Ans: Frequency of oscillation = 3.52 Hz and amplitude = 1 cm

Example – 03:

A mass of 250 g is suspended from a spring of constant 9 N/m. The mass is pulled 10 cm from its equilibrium position and released. Find its speed when it crosses the equilibrium position.

Given: Force constant = 9 N/m, mass attached = m = 250 g = 0.250 kg, distance through which the mass is pulled down = y =  10 cm = 0.1 m

To Find: The speed at equilibrium = v = ?

Solution:

the amplitude = distance through which mass is pulled down.

amplitude = a = 10 cm = 0.1 m

The speed at equilibrium is maximum

v_max = ωa = 6 x 0.1 = 0.6 m/s

Ans: The speed at equilibrium is 0.6 m/s

Example – 04:

One end of steel spiral spring of length 8 cm is fixed to a rigid support. When a mass is suspended from the free end, its length becomes 14 cm. Calculate the periodic time of oscillations of the mass, if displaced vertically.

Given: Original length = 8cm, final length = 14 cm, extension in spring = l = 14 – 8 = 6 cm = 0.06 m,

To Find: Periodic time = T = ?

Solution:

Let m be the mass attached to spring

Force constant k = mg/l = mg / 0.06  = 100mg/6 N/m

Ans: Period of oscillation = 0.4917 s

Example – 05:

A mass M attached to a light spring oscillates with a period of 2 s. If the mass is increased by 2 Kg, the period increases by 1 s. Find M.

Given: m₁ = M, T₁ = 2 s, m₂ = M + 2, T₂ = 2 + 1 = 3 s

To Find: M = ?

Solution:

Let  k be the force constant of the spring

∴ 9M = 4M + 8

∴ 5M = 8

∴  M = 8/5 = 1.6 kg

Ans: The value of M is 1.6 kg

Example – 06:

Body A and a body B have the same mass. The body A suspended from a light spring of constant k₁ and body B is suspended from a light spring of constant k₂. The bodies are set into oscillations vertically in such a way that their maximum velocities are equal. Find the ratio of the amplitudes of vibration of the bodies.

Their maximum velocities are equal

Ans: The ratio of amplitude = (k₂/k₁)^1/2.

Example – 07:

A body is supported by a spiral spring and causes a stretch of 1.5 cm is the spring. If the mass is now set into vertical oscillations of small amplitudes what is the periodic time of oscillations? g = 9.8 m/s².

Given: Elongation in wire = l = 1.5 cm, g = 9.8 m/s².

To Find: Period of oscillation = T =?

Solution:

Let m be the mass attached to the spring

Ans: Period of oscillation is 0.246 s.

Example – 08:

A spring elongates 2 cm when stretched by a load by 80 g. A body of mass 0.6 kg is attached to the spring and then displaced through 8 cm from its equilibrium position. Calculate the energy of the system in the position. Also, determine the speed of the body when it is 4 cm from the equilibrium position.

Given: Elongation in length of spring = l = 2 cm , mass attached = m = 80 g, g = 9.8 m/s². Displacement of mass = 8 cm = 0.08 m

To Find: Energy of system = ? Speed at x = 4 cm

Solution:

Force constant k = F/l = 80 x 980 / 2 = 3.92 x 10⁴ dyne/cm

Energy of system = 1/2m ω²a² = ½ x 80 x (22.14)²(8)²

Energy of system = 1.254 x 10⁶ erg = 0.1254 x 10⁷ erg = 0.1254 J

∴ v = 153.4 cm/s = 1.53 m/s

Ans: Energy of the system = 0.1254 J and speed at the displacement of 4 cm is 1.53 m/s

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Science > Physics > Oscillations: Simple Harmonic Motion > The Vibration of Vertical Spring

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In this article, we shall study the composition of two SHM. Sometimes particle is acted upon by two or more linear SHMs. In such a case, the resultant motion of the body depends on the periods, paths and the relative phase angles of the different SHMs to which it is subjected.

Consider two SHMs having same period and parallel to each other, where a1 and a2 are amplitudes of two SHMs respectively. a1 anda2 are initial phase angle of two SHMs respectively, whose displacements are given by

x₁ = a₁ Sin (ωt + α₁)   and x₂ = a₂ Sin (ωt + α₂)

Resultant displacement of the particle subjected to above SHMs is given by

x = x₁ + x₂

∴ x  = a₁ Sin (ωt + α₁)  +  a₂ Sin (ωt + α₂)

∴   x = a₁ [Sinωt . Cosα₁ + Cosωt . Sinα₁] + a₂ [Sinωt . Cosα₂ + Cosωt . Sinα₂]

∴   x = a₁ Sinωt . Cosα₁ + a₁ Cosωt . Sinα₁ + a₂ Sinωt . Cosα₂ + a₂ Cosωt . Sinα₂

∴   x = a₁ Sinωt . Cosα₁ + a₂ Sinωt . Cosα₂ + a₁ Cosωt . Sinα₁ + a₂ Cosωt . Sinα₂

∴   x = Sinωt .(a₁  Cosα₁  + a₂ Cosα₂) + Cosωt . (a₁ Sinα₁ + a₂  Sinα₂) ………….. (1)

Let, (a₁ Cosα₁  + a₂ Cosα₂)   = R Cos δ … (2)

(a₁ Sinα₁ + a₂ Sinα₂) = R Sin δ    ……(3)

From Equations (1), (2) and (3)

x = Sin ωt (R Cos δ)   + Cos ωt (R Sin δ)

∴   x = R (Sin ωt Cos δ   + Cos ωt Sin δ)

∴ x = R Sin (ωt + δ)  ………..(4)

Equation (4) indicates that resultant motion is also a S.H.M. along the same straight line as two parent SHMs and of the same period and initial phase δ .

Squaring equations (2) and (3) and adding them

(R Cos δ)²+    (R Sin δ)² =   (a₁  Cosα₁  + a₂ Cosα₂)² +   ( a₁ Sinα₁ + a₂  Sinα₂ )²

∴   R² Cos² δ+    R² Sin² δ =    a₁² Cos² α₁ + a₂² Cos² α₂ +2 a₁ a₂ Cos α₁ Cos α₂

+ a₁² Sin² α₁ + a₂²Sin²α₂ + 2 a₁ a₂ Sin α₁ Sin α₂

∴   R² (Cos² δ + Sin² δ) =    a₁² (Cos² α₁ + Sin² α₁) + a₂² (Cos² α₂ + Sin²α₂)

+2 a₁ a₂ (Cos α₁ Cos α₂ +Sin α₁ Sin α₂)

∴   R² (1) =    a₁² (1) + a₂² (1) +2 a₁ a₂ Cos (α₁ – α₂)

∴   R² =    a₁² + a₂² +2 a₁ a₂ Cos (α₁ – α₂)

Dividing equation (3) by (2)

From Equations (6) and (7) we can find the resultant and initial phase of resultant S.H.M.

Special Cases:

Case 1: When the two SHMs are in the same phase then (α₁ – α₂) =  0

If the two SHMs have the same amplitude then, a₁ = a₂ = a

∴ R   =  a + a   =  2a

Case 2: When the two SHMs are in opposite phase then, (α₁ – α₂) = π

If the two SHMs have the same amplitudes then, a₁ = a₂ = a

R   = a – a = 0

Case 3: When the phase difference is (α₁ – α₂)   = π / 2

If the two SHMs have the same amplitude then, a₁ = a₂ = a

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Science > Physics > Oscillations: Simple Harmonic Motion > Composition of Two SHM

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In this article, we shall study to solve numerical problems to calculate potential energy, kinetic energy, and total energy of particle performing S.H.M.

Example – 01:

A particle of mass 10 g performs S.H. M. of amplitude 10 cm and period 2π s. Determine its kinetic and potential energies when it is at a distance of 8 cm from its equilibrium position.

Given: Mass = m = 10 g, amplitude = a = 10 cm, Period = T = 2π s, displacement = x = 8 cm

To Find: Kinetic energy =? and Potential energy = ?

Solution:

Angular velocity ω = 2π/T = 2π/2π = 1 rad/s

Kinetic energy = 1/2 mω² (a² – x²) =1/2 x 10 x 1²(10² – 8²)

∴ Kinetic energy = 5 x (36) = 180 erg = 1.8 x 10^-5 J

Potential energy = 1/2 mω²x²

∴ Potential energy = 1/2 x 10 x 1² x 8² = 320 erg = 3.2 x 10^-5 J

Ans: Kinetic energy = 1.8 x 10^-5 J and potential energy = 3.2 x 10^-5 J

Example – 02:

A particle of mass 10 g executes linear S.H.M. of amplitude 5 cm with a period of 2 s. Find its PE and KE, 1/6 s after it has crossed the mean position.

Given: Mass = m = 10 g, amplitude = a = 5 cm, Period = T = 2 s, time elapsed = 1/6 s, particle passes through mean position, α = 0.

To Find: Kinetic energy =? and Potential energy =?

Solution:

Angular velocity ω = 2π/T = 2π/2 = π rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴ x = 5 sin (π x 1/6 + 0)

∴ x = x = 5 sin (π/6) = 5 x 1/2 = 2.5 cm

Kinetic energy = 1/2 mω² (a² – x²) =1/2 x 10 x π² (5² – 2.5²)

∴ Kinetic energy = 5 x 3.142²(25- 6.25) = 5 x 3.142²(18.75)

∴ Kinetic energy = 925.5 erg = 9.26 x 10^-5 J

Potential energy = 1/2 mω²x²

∴ Potential energy  = 1/2 x 10 x π² x 2.5² = 5 x 3.142² x 2.5²

= 308.5 erg = 3.09 x 10^-5 J

Ans: Kinetic energy = 9.26 x 10^-5 J and potential energy = 3.09 x 10^-5 J

Example – 03:

The total energy of a particle of mass 0.5 kg performing S.H.M. is 25 J. What is its speed when crossing the centre of its path?

Given: Mass = m = 0.5 kg, Total energy T.E. = 25 J

To Find: Maximum speed = v_max =?

Solution:

The speed when crossing mean position is a maximum speed

Total energy = 1/2 mω²a²

∴ 25 = 1/2 x 0.5 x ω²a²

∴ ω²a² = 25 x 2/ 0.5 = 100

∴ ωa = 10 m/s

But ωa = v_max = 10 m/s

Ans: The speed when crossing mean position is 10m/s

Example – 04:

A particle performs a linear S.H.M. of amplitude 10 cm. Find at what distance from the mean position its PE is equal to its KE.

Given: P.E. = K.E.

To Find: Distance = x=?

Solution:

P.E. = K.E.

∴ 1/2 mω²x² = 1/2 mω² (a² – x²)

∴ x² =  a² – x²

∴ 2x² = a²

∴ x = ± a/√2 = ±10/√2 = ±5√2 cm

Ans:  At a distance of 5√2 cm from either side of the mean position K.E. = P.E.

Example – 05:

Find the relation between amplitude and displacement at the instant when the K.E. of a particle performing S.H. M. is three times its P.E.

Given: K.E. = 3 x P.E.

To Find: Distance = x=?

Solution:

K.E. = 3 x P.E.

∴ 1/2 mω² (a² – x²)  = 3 x 1/2 mω²x² 

∴ a² – x² = 3x² 

∴ 4x² = a²

∴ x = ± a/2, where a = amplitude

Ans:  At a distance of a/2 cm from either side of the mean position K.E. = 3 x P.E.

Example – 06:

When is the displacement in S.H.M. one-third of the amplitude, what fraction of total energy is kinetic and what fraction is potential? At what displacement is the energy half kinetic and half potential?

Part – I:

Given: x = a/3

To Find: K.E/T.E. =? and P.E./T.E. =?

Solution:

Part – II

Given: P.E. = K.E.

To Find: Distance = x =?

Solution:

P.E. = K.E.

∴ 1/2 mω²x² = 1/2 mω²(a² – x²)

∴ x²  =  a² – x²

∴ 2x² = a²

∴ x = ± a/√2

Ans:  The fraction of K.E = 8/9, fraction of P.E. = 1/9, required displacement = ± a/√2 unit

Example – 07:

An object of mass 0.2 kg executes S.H.M. along the X-axis with a frequency of 25 Hz. At the position x = 0.04 m, the object has a K.E. of 0.5 J and P.E. of 0.4 J. Find the amplitude of its oscillations.

Given: Mass = m = 0.2 kg, frequency = n = 25 Hz, displacement = x = 0.04 m = 4 cm, K.E. = 0.5 J, P.E. = 0.4 J

To Find: Amplitude = a =?

Solution:

Angular speed ω = 2πn = 2 x π x 25 = 50π rad/s

∴   5x² = 4a² – 4x²

∴   9x² = 4a²

∴ 4a² = 9x 4² = 144

∴ a² = 36

∴ a = 6 cm

Ans: The amplitude = 6 cm

Example – 08:

The amplitude of a particle in S.H. M. is 2 cm and the total energy of its oscillation is 3 x 10^-7 J. At what distance from the mean position will the particle be acted upon by a force of 2.25 x 10^-5 N when vibrating?

Given: amplitude = a = 2 cm, Total energy = T.E. = 3 x10^-7 J = 3 x10^-7 x 10⁷ = 3 erg, Force = 2.25 x 10^-5 N = 2.25 x 10^-5 x 10⁵ = 2.25 dyne

To Find: Distance = x =?

Solution:

T.E =1/2 mω²a²

∴ 3 =1/2 mω²(2)²

∴ mω² =3/2 ………… (1)

Now Force F = mf = mω²x

∴ 2.25 = (3/2)x

∴ x = 2.25 x 2 /3 = 1.5 cm

Ans: At a distance of 1.5 cm from the mean position will the particle be acted upon by a force of 2.25 x 10^-5 N

Example – 09:

A body of mass 100 g performs S.H.M. along a path of length 20 cm and with a period of 4 s. Find the restoring force acting upon it at a displacement of 3 cm from the mean position? Find also the total energy of the body.

Given: mass = m = 20 g, Path length = 20 cm, amplitude = a = 20/2 = 10 cm, Period = T = 4s,

To Find: Restoring force = F =? Total energy = T.E. = ?

Solution:

Angular speed ω = 2π/T = 2π/4 = π/2 rad/s

Restoring force F = mf = mω²x

F = 100 x (π/2)² x 3 = 740.4 dyne = 740.4 x 10^-5 N = 7.404 x 10^-3 N

T.E. = 1/2 x 100 x (π/2)²x 10² = 1.234 x 10⁴ erg

T.E. = 1.234 x 10⁴ x 10^-7 J = 1.234 x 10^-3 J

Ans: Restoring force = 7.404 x 10^-3 N; total energy = 1.234 x 10^-3 J

Example – 10:

A particle of mass 200 g performs S.H.M. of amplitude 0.1m and period 3.14 second. Find its K.E. and P.E. when it is at a distance of 0.03 m from the mean position.

Given: mass = m = 200 g, amplitude = a = 0.1 m = 10 cm, period = T = 3.14 s, Distance = x = 0.03 m = 3 cm,

To Find: K.E. =? and P.E. = ?

Solution:

Angular speed ω = 2π/T = 2π/3.14 = 2 rad/s

Kinetic energy = 1/2 mω²(a² – x²) =1/2 x 200 x 2²(10² – 3²)

∴ Kinetic energy = 100 x 4 x (100 -9) = 3.64 x 10⁴ erg

∴ Kinetic energy = 3.64 x 10⁴ x 10^-7 J = 3.64 x 10^-3 J

Potential energy = 1/2 mω²x²

∴ Potential energy = 1/2 x 200 x 2² x 3² = 3.6 x 10³ J

∴ Potential energy = 3.6 x 10³ x 10^-7 = 3.6 x 10^-4 J

Ans: K.E. = 3.64 x 10^-3 J; P.E. = 3.6 x 10^-4 J

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In this article, we shall study the concept and expression of the total energy of a particle performing S.H.M. and its constituents.

Kintetic Energy of Particle Performing Linear S.H.M.:

Consider a particle of mass ‘m’ which is performing linear S.H.M. of amplitude ‘a’ along straight line AB, with the centre O.  Let the position of the particle at some instant be at C, at a distance x from O.

This is an expression for the kinetic energy of particle S.H.M.

Thus the kinetic energy of the particle performing linear S.H.M. and at a distance of x₁ from the mean position is given by

Special cases:

Case 1: Mean Position:

The kinetic energy of particle performing S.H.M. is given by

For mean position x₁ = 0

Case 2: Extreme position:

The kinetic energy of particle performing S.H.M. is given by

For mean position x₁ = a

Potential Energy of Particle Performing Linear S.H.M.:

Consider a particle of mass ‘m’ which is performing linear S.H.M. of amplitude ‘a’ along straight line AB, with the centre O.  Let the position of the particle at some instant be at C, at a distance x from O.

Particle at C is acted upon by restoring force which is given by F = – mω²x

The negative sign indicates that force is restoring force.

Let. External force F’ which is equal in magnitude and opposite to restoring force acts on the particle due to which the particle moves away from the mean position by small distance ‘dx’ as shown. Then

F’ = mω²x

Then the work done by force F’ is given by

dW =  F’ . dx

dW = mω²x dx

The work done in moving the particle from position ‘O’ to ‘C’ can be calculated by integrating the above equation

This work will be stored in the particle as potential energy

This is an expression for the potential energy of particle performing S.H.M.

Special cases:

Case 1: Mean Position:

The potential energy of particle performing S.H.M. is given by

For mean position x₁ = 0

∴ E_P = 0

Case 2: Extreme position:

The potential energy of particle performing S.H.M. is given by

For mean position x₁ = a

Total Energy of Particle Performing Linear S.H.M.:

The Kinetic energy of particle performing S.H.M. at a displacement of x₁ from the mean position is given by

The potential energy of particle performing S.H.M. at a displacement of x₁ from mean position is given by

The total energy of particle performing S.H.M. at a displacement of x₁ from the mean position is given by

Since for a given S.H.M., the mass of body m, angular speed ω and amplitude a are constant, Hence the total energy of a particle performing S.H.M. at C is constant i.e. the total energy of a linear harmonic oscillator is conserved. It is the same at all positions. The total energy of a linear harmonic oscillator is directly proportional to the square of its amplitude.

Variation of Kinetic Energy and Potential Energy in S.H.M Graphically:

Relation Between the Total Energy of particle and Frequency of S.H.M.: 

The quantities in the bracket are constant. Therefore, the total energy of a linear harmonic oscillator is directly proportional to the square of its frequency.

Relation Between the Total Energy and Period of S.H.M.: 

The quantities in the bracket are constant. Therefore, the total energy of a linear harmonic oscillator is inversely proportional to the square of its period.

Expressions for Potential Energy, Kinetic Energy and Total Energy of a Particle Performing S.H.M. in Terms of Force Constant:

Potential energy: 

The potential energy of particle performing S.H.M. is given by

This is an expression for the potential energy of particle performing S.H.M. in terms of force constant.

Kinetic energy: 

The kinetic energy of particle performing S.H.M. is given by

This is an expression for Kinetic energy of particle performing S.H.M. in terms of force constant.

Total energy: 

The total energy of particle performing S.H.M. is given by

This is an expression for the total energy of particle performing S.H.M. in terms of force constant.

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Science > Physics > Oscillations: Simple Harmonic Motion > The Energy of Particle Performing S.H.M.

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In this article, we shall study graphical representation of S.H.M. i.e. variation in displacement, velocity, and acceleration with time for a body performing S.H.M. starting from a) the mean position and b) from the extreme position.

Graphical Representation of Linear S.H.M. of a Particle Starting from Mean Position:

The general equation for the displacement of a particle performing linear S.H.M. at any instant ‘t’ is given by

x = a  sin (ωt + α )

Where a = amplitude of S.H.M., ω = angular speed of S.H.M.,

α = Initial phase of S.H.M.

As particle is starting from mean position, α = 0

x  =  a  sin ωt   …….. (1)

Velocity of particle performing S.H.M.can be obtained by differentiating above expression

v = dx/dt = a cos ωt . ω = ωa cos ωt

v =  ωa cos ωt   …….. (2)

Acceleration of particle performing S.H.M. can be obtained by differentiating above expression

f = dv/dt = ωa (-sin ωt)  ω

f = dv/dt = – ω²a sin ωt    …….. (3)

From equation (1) and (3) we have

f = dv/dt = – ω²x    …….. (4)

Using equations (1), (2) and (4) and knowing ω = 2π/T we prepare following table

The graphs of displacement, velocity and acceleration versus time are as follows:

Graphical Representation of Linear S.H.M. of a Particle Starting from Extreme Position:

The general equation for displacement of a particle performing linear S.H.M. at any instant ‘t’ is given by

x = a  sin (ωt + α )

Where a = amplitude of S.H.M., ω = angular speed of S.H.M., α = Initial phase of S.H.M.

As particle is starting from mean position, α = π/2

x = a  sin (ωt + π/2 )

x  =  a  cos ωt   …….. (1)

Velocity of particle performing S.H.M.can be obtained by differentiating above expression

v = dx/dt = a (- sin ωt) . ω = – ωa sin ωt

v =  – ωa sin ωt …….. (2)

Acceleration of particle performing S.H.M. can be obtained by differentiating above expression

f = dv/dt = – ωa (cos ωt)  ω

f = dv/dt = – ω²a cos ωt    …….. (3)

From equation (1) and (3) we have

f = dv/dt = – ω²x    …….. (4)

Using equations (1), (2) and (4) and knowing ω = 2π/T we prepare following table

The graphs of displacement, velocity and acceleration versus time are as follows:

Conclusions:

• Graphs are drawn for displacement, velocity and acceleration against time and curves are obtained as shown.  As the curves have the shape same as the sine curve, the curves are called as harmonic curves.
• From the graph, we can conclude that the displacement, velocity, and acceleration are the periodic functions of time.
• From the graph, we can see that velocity is 90° (π/2 radians) out of phase with displacement, whereas acceleration is 180° (π radians) out of phase with displacement. Similarly, acceleration is 90° (π/2 radians) out of phase with velocity.
• The velocity leads the displacement by a phase difference of π/2 radians.
• The acceleration lags behind displacement by a phase of π radians.
• The displacement and acceleration are maximum at the extreme position while velocity is minimum at the same position. Similarly, the displacement and acceleration are minimum at the mean position while velocity is maximum at the same position.
• All curves repeat after a phase of 2π radians.

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Science > Physics > Oscillations: Simple Harmonic Motion > Graphical Representation of S.H.M.

The post Graphical Representation of S.H.M. appeared first on The Fact Factor.

For uniform circular motion Angular speed ω, Linear speed t, kinetic energy, Angular momentum (L) are constant. The angular acceleration α and the tangential acceleration a_T are zero.

Problems Based on Hands of Clock:

1. The second hand of a clock takes 60 seconds to complete one rotation. Its angular speed is 0.105 rad /s.
2. The minute hand of a clock takes 60 minutes = 60 x 60 seconds to complete one rotation. Its angular speed is 1.746 x 10^-3 rad /s.
3. The hour hand of a clock takes 12 hours = 12 x 60 x 60 seconds to complete one rotation. Its angular speed is 1.455 x 10^-4 rad /s.
4. The ratio of angular speeds of the second hand of a clock and the minute hand of a clock is 60:1.
5. The ratio of angular speeds of the minute hand of a clock and an hour hand of a clock is 12:1.
6. The ratio of angular speeds of the second hand of a clock and the hour hand of a clock is 720:1.

Example – 01:

Calculate the angular speed of the second hand, minute hand and hour hand of a clock.

Given: For second Hand T_S = 60 sec, For minute hand T_M = 60 min = 60 x 60 sec, For hour hand T_H = 12 hr = 12 x 60 x 60 sec.

To Find: Angular speed ω_S=? ω_M=? ω_H=?

Solution: 

The tips of the second hand, the minute hand, and the hour hand perform the uniform circular motion.

Ans: Angular speed of second hand = 0.105 rad/s,

Angular speed of minute hand =1.746 x 10^-3 rad/s,

Angular speed of hour hand =1.454 x 10^-4 rad/s.

Example – 02:

What is the angular velocity of the minute hand of a clock? What is the angular displacement of the minute hand in 20 minutes? If the minute hand is 5 cm long, what is the linear velocity of its tip?

Given: For minute hand  T_M = 60 min = 60 x 60 sec, t = 20 min = 20 x 60 sec, r = 5 cm = 5 x 10^-2 m

To Find: Angular speed ω_M=?, Angular displacement  θ =?, Linear velocity = v =?

Solution: 

The tip of the minute hand performs the uniform circular motion.

Ans: The angular speed of minute hand =1.746 x 10^-3 rad/s,

The angular displacement of minute hand in 20 minutes =2.095 rad,

The linear speed of the tip of minute hand = 8.73 x 10^-5 m/s.

Example – 03:

What is the angular displacement of the minute hand of a clock in 25 minutes?

Given: For minute hand  T_M = 60 min = 60 x 60 sec, t = 25 min = 25 x 60 sec,

To Find:  Angular displacement  θ =?

Solution: 

The tip of the minute hand performs the uniform circular motion.

Ans: Angular displacement of minute hand = 2.618 rad

Example – 04:

What is the angular velocity of the second hand of a clock? If the second hand is 10 cm long find the linear velocity of its tip.

Given: For second hand  T_s = 60  sec, r = 10 cm = 10 x 10^-2 m,

To Find:  Linear speed = v =?

Solution: 

The tip of the second hand performs the uniform circular motion.

Now v = r ω = 10 x 10^-2 x 0.105 = 1.05 x 10^-2 m/s = 1.05 cm/s

Ans: Angular velocity of second hand = 0.105 rad/s, Linear speed of tip of second hand = 1.05 cm/s

Example – 05:

The second hand of a watch is 1.5 cm long. Find the linear speed of a point on the second hand at a distance of 0.5 cm from the tip.

Given: For second hand  T_s = 60  sec, r = 1.5 cm – 0.5 cm = 1 cm = 1 x 10^-2 m,

To Find:  Linear speed = v =?

Solution:

The tip of the second hand performs the uniform circular motion.

Now v = r ω = 1 x 10^-2 x 0.105 = 1.05 x 10^-3 m/s = 1.05 mm/s

Ans: Linear speed of the point on the second hand = 1.05 mm/s

Example – 06:

The extremity of the hour hand of a clock moves 1/20th as fast as the minute hand. What is the length of the hour hand if the minute hand is 10 cm long?

Given: v_H = 1/20 v_M , r_M = 10 cm

To Find: r_H =?

Solution:

Ans: Length of hour hand = 6 cm

Example – 07:

The length of an hour hand of a wristwatch is 1.5 cm. Find the magnitudes of following w.r.t. tip of the hour hand a) angular velocity b) linear velocity c) angular acceleration d) radial acceleration e) tangential acceleration f) linear acceleration

Given: r = 1.5 cm = 1.5 x 10^-2 m, For hour hand T_H = 12 hr = 12 x 60 x 60 sec

To Find: Angular velocity = ω = ?, linear velocity = v = ?, angular acceleration = α = ?, radial acceleration = a_r = ?, tangential acceleration = a_T =?, linear acceleration a = ?

Solution:

Now v = r ω = 1.5 x 10^-2 x 1.454 x 10^-4 = 2.181 x 10^-6 m/s

Tip of hour hand performs uniform circular motion

∴  Angular acceleration = α = 0 and tangential acceleartion = a_T = 0

Radial acceleration is given by

a_r = v² /r = ( 2.181 x 10^-6 ) ² / (1.5 x 10^-2) = 3.171 x 10^-10 m/s²

a² = a_r ² + a_T ²

as a_T = 0, a = a_r = 3.171 x 10^-10 m/s²

Ans: Angular speed = 1.454x 10^-4 rad/s, Linear speed =2.181 x 10^-6 m/s,
Angular acceleration = 0, Radial acceleration = 3.171 x 10^-10 m/s².
Tangential acceleration = 0, Linear acceleartion = 3.171 x 10^-10 m/s².

Problems Based on Earth:

The earth takes 24 hours to complete one rotation about its axis. The angular speed of the earth of its rotation about its axis is 7.273 x 10^-5 rad /s.

Example – 08:

Calculate the angular velocity of the earth due to its spin motion.

Given: For the earth = T = 24 hr = 24 x 60 x 60 sec

To Find: Angular velocity = ω = ?

Solution:

Ans: Angular speed of the earth due to its spin motion is 7.273 x 10^-5 rad/s.

Problems Based on Rotating Discs:

Example – 09:

A turntable rotates at 100 rev/sec. Calculate its angular speed in rad/s and degrees/s.

Given: n = 100 r.p.s.

To Find: Angular speed =?

Solution:

Ans: Angular speed = 10.47 rad/s, Angular speed = 600 degrees/s

Example – 10:

Propeller blades of an aeroplane are 2 m long. When the propeller is rotating at 1800 rev/min, compute the tangential velocity of the tip of the blade. Also, find the tangential velocity at a point on the blade midway between tips and axis.

Given: r = 2 m, Angular speed = N = 1800 r.p.m.

To Find: v_Tip =? v_Mid =?

Solution:

Ans: The angular speed of the tip of blade = 376.8 rad/s, AThe angular speed of point midway = 188.4 rad/s

Example – 11:

A turntable has a constant angular speed of 45 r.p.m. Express this in rad per second and degrees per second. If the radius of the turntable is 0.5 m, what is the linear speed of a point on the rim?

Given: N = 45 r.p.m., r = 0.5 m

To Find: angular speed in rad/s and degrees/s, linear velocity = v = ?

Solution:

Ans: Angular speed = 2.355 rad/s = 270 degrees/s Linear speed on point on rim = 2.355 m/s.

Example – 12:

The linear velocity of a point on the rotating disc is 3 times greater than at a point on the at a distance of 8 cm from it. What is the diameter of the disc?

Given: v_T = 3 v_P, Let r_T = r, r_P = (r – 8) cm

To Find: Diameter of the disc =?

Solution:

Angular velocity for both the points is the same.
Ans: Diameter of disc = 24 cm.

Example – 13:

A disc has a diameter of one metre and rotates about an axis passing through its centre and at right angles to its plane at the rate of 120 rev/min. What is the angular and linear speed of a point on the rim and at a point halfway to the centre.

Given: d = 1m , r_T = r = 0.5 m, N = 120 r.p.m., for r_P = r/2 = 0.25 m

To Find: v_T = ?, V_P = ?

Solution:

Angular velocity for both the points is the same.

The linear speed of point on the rim

v_T =  r_Tω  = 0.5 x 12.56 = 6.28 m/s

v_P =  r_Pω  = 0.25 x 12.56 = 3.14 m/s

Ans: Angular speed of point on rim = 12.57 rad/s, Linear peed of point on rim = 6.28 m/s,
Angular speed of point on halfway =12.57m/s, Linear speed of point on halfway = 3.14 m/s.

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Science > Physics > Circular Motion > Numerical Problems: Circular Motion

The post Numerical Problems: Circular Motion appeared first on The Fact Factor.

Example – 1:

a particle executing simple harmonic motion has a period of 6 s and its maximum velocity during oscillations is 6.28 cm/s. Find the time taken by it to describe a distance of 3 cm from its equilibrium position.

Given: Period = T = 6 s, V_max = 6.28 cm/s, x = 3 cm, particle passes through mean position, α = 0.

To Find: Time taken = t =?

Solution:

Angular velocity = ω = 2π/T = 2π/6  = π/3 rad/s

v_max = ωa

∴  a = v_max/ω  = 6.28 /(π/3) = 6 cm

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  3 = 6 sin ((π/3)t + 0)

∴  3/6 = sin ((π/3)t)

∴  (π/3)t = sin^-1(1/2) = π/6

∴  t = 1/2 s = 0.5 s

Ans: Time taken = 0.5 s

Example – 2:

The maximum velocity of a particle performing simple harmonic motion is 6.28 cm/s. If the length of its path is 8 cm, calculate its period.

Given: path length = 8 cm, amplitude = 8/2 = 4 cm, V_max = 6.28 cm/s,

To Find: Period = T =?

Solution:

v_max = ωa

∴  ω = v_max/a  = 6.28/4 = 1.57 rad/s

T = 2π /ω = (2 x 3.14)/ 1.57 = 4 s

Ans: Period = 4 s

Example – 3

A particle performs simple harmonic motion of amplitude 3 cm. If its acceleration in the extreme position is 27 cm/s², find the period.

Given: Amplitude = a = 3 cm, acceleration at extreme position = f = 27 cm/s²,

To Find: Period = T =?

Solution:

At extreme position acceleration is maximum, f_max = 27 cm/s²

f_max = ω²a

∴  ω² = f_max/a  = 27/3 = 9

∴  ω  = 3 rad/s

T = 2π /ω = (2 x 3.14)/ 3 = 2.09 s

Ans: Period = 2.09 s

Example – 4:

A particle executing S.H.M. has a maximum velocity of 0.16 cm/s and a maximum acceleration of 0.64 m/s². Calculate its amplitude and the period of oscillations.

Given: vmax = 0.16 cm/s, f max = 0.64  m/s².

To Find: Amplitude = a =? and Period = T = ?

Solution:

f_max = ω²a ………. (1)

v_max = ωa  ………. (2)

Dividing equation (1) by (2)

f_max /v_max  = ω

∴  ω = 0.64/0.16 = 4 rad/s

Substituting in equation (2)

0.16 = 4 x a

∴ a = 0.04 cm

T = 2π /ω = (2 x 3.14)/ 4 = 1.57 s

Ans: amplitude = 0.04 cm and period = 1.57 s

Example – 5:

A block is on a piston which is moving vertically up and down with simple harmonic motion of period one second. At what amplitude of motion will the block and piston separate? At which point in the path of motion will the separation take place?

Given: Period = T = 1s

To Find: amplitude = a = ?

Solution:                      

Angular velocity = ω = 2π/T = 2π/1 =  2π rad/s

At the topmost point, the block and piston will separate.

At topmost point acceleration is maximum. Hence force is maximum

Maximum force on the block = weight of the block

m. f_max =  mg

∴   f_max =  g

∴   ω²a =  g

∴   a =  g / ω²  = 980/ (2 x 3.142)² = 24.82 cm

Ans: At amplitude = 24.82 cm block will separate at the topmost point of the path

Example – 6:

A particle performs simple harmonic motion with a period of 12 s. If its velocity is 6 cm/s two seconds after crossing the mean position, what is the amplitude of its motion?

Given: Period = T = 12 s, v = 6 cm/s, time elapsed = t = 2 s, particle passes through mean position, α = 0.

To Find: amplitude = a =?

Solution:

Angular velocity = ω = 2π/T = 2π/12 = π/6 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  x = a sin ( π/6 x 2 + 0)

∴  x = a sin ( π/3) = a √3/2   cm

Ans: The amplitude of motion is 22.92 cm

Example – 7:

A particle in simple harmonic motion has a velocity of 10 cm/s when it crosses the mean position. If the amplitude of its oscillations is 2 cm, find the velocity. When it is midway between the mean and extreme positions.

Given: Velocity at mean position = v_max = 10 cm/s, amplitude = a = 2 cm, Displacement midway between the mean and extreme positions, hence x = a/2 = 2/2 = 1 cm.

To Find: Velocity = v =?

Solution: 

We have v_max = ωa

∴  10 = ω x 2

∴  ω = 10/2 = 5 rad/s

Ans: velocity at midway between the mean and extreme positions is 8.66 cm/s

Example – 8:

Show that the velocity of a particle performing simple harmonic motion is half the maximum velocity at a displacement of √3/2 times its amplitude.

Given: Displacement x = a√3/2

To Show: v = 1/2 v_max.

Solution:

Example – 9:

A particle performs S.H.M. of amplitude 10 cm. Its maximum velocity during oscillations is 100 cm/s. What is its displacement when the velocity is 60 cm/s?

Given: amplitude = 10 cm, V_max = 100 cm/s, v = 60 cm/s

To Find: displacement = x =?

Solution:

v_max = ωa

∴  ω = v_max/a  = 100/10 = 10 rad/s

Ans: Displacement = 8 cm

Example – 10:

A particle performing S.H.M. along a straight line has a velocity of 4π cm/s when its displacement is √12 cm. If the maximum acceleration it can attain is 16π² cm/s², find the amplitude and the period of its oscillations.

Given: vmax = 4π cm/s, f max = 16π² m/s² , Displacement = √12 cm

To Find: Amplitude = a =? and Period = T = ?

Solution:

f_max = ω²a

∴  16π²  = ω²a  ………. (2)

From equations (1) and (2) we have

ω²(a² – 12) =  ω²a

∴  (a² – 12) =  a

∴  a² – 12 – a = 0

∴ (a  – 4)(a + 3) = 0

∴  a = 4 cm or a = – 3 cm

Amplitude is maximum displacement hence a = 3 cm <  √12 cm is not possible.

∴  a = 4 cm

substituting in equation (2)

16π²  = ω²(4)

∴   ω² = 4π²

∴   ω = 2π rad/s

∴   T = 2π /ω = 2π /2π =  1 s

Ans: amplitude = 4 cm and period = 1 s

Example – 11

A particle of mass of 10 g performs S.H.M. of period 5 s and has an amplitude of 8 cm. Find its velocity when it is at a distance of 6 cm from the equilibrium position. Find also the maximum velocity and maximum force acting on it.

Given: mass = m = 10 g, Period = T = 5 s, amplitude = a = 8 cm, displacement = x = 6 cm, particle passes through mean position, α = 0.

To Find: velocity = v = ?, v_max = ?, F_max = ?

Solution:

Angular velocity = ω = 2π/T = 2π/5  rad/s

∴   v_max = ωa  = 2π/5 x 8 = 10.05 cm/s

f_max = ω²a = ( 2π/5)² x 8  = 12.63 cm/s²

F_max = m. f_max = 10 x 12.63 = 126.3 dyne

∴   F_max =  126.3 x 10^-5 N = 1.263 x 10^-3 N

Ans: velocity =6 .65 cm/s;  maximum velocity =10.05 cm/s;  maximum force = 1.263 x 10^-3 N

Example – 12:

If a particle performing S.H. M. starts from the extreme position after an elapse of what fraction of the period will the velocity of the particle be half the maximum velocity?

Given: v = 1/2 v_max.   particle starts from extreme position, α = π/2.

Fo Find: Fraction of time = t/T =?

Solution:

∴  4a² – 4x² = a²

∴   4x² = 3a²

∴   2x = a√3

∴  x = a√3/2

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  a√3/2 = 1 sin ((2π/T)t + π/2)

∴  3/2 = cos ((2π/T)t)

∴  (2π/T)t = cos^-1(3/2) = π/6

∴  t /T = 1/12 s

Ans: fraction of the period is 1/12 s

Example – 13:

A particle performs a linear S.H.M. Its velocity is 3 cm/s when it is at 4 cm from the mean position and 4 cm/s when it is at 3 cm from the mean position. Find the amplitude and the period of S.H.M.

Given: v₁ = 3 cm/s at x₁ = 4cm and v₂ = 4 cm/s at x₂ = 3cm

To Find: Amplitude = a =? Period = T=?

Solution:

∴  16a² -256 = 9a² -81

∴  16a² – 9a²  = 256  – 81

∴  7a²  = 175

∴  a²  = 25

∴  a = 5

Now T = 2π/ω = 2 x 3.14 /1 = 6. 28 s

Ans: Amplitude = 5 cm and period = 6.28 s

Example – 14

The velocities of a particle performing linear S.H.M. are 0.13 m/s and 0.12 m/s when it is at 0.12 m and 0.13 m respectively from the mean position. Find its period and amplitude.

Given: v₁ = 0.13 m/s = 13 cm/s at x₁ = 0.12 m = 12 cm and v₂ = 0.12 m/s = 12 cm/s at x₂ = 0.13 m = 13 cm

To Find: Amplitude = a =? Period = T=?

Solution:

∴  144a² – 144 x 144 = 169a² – 169x 169

∴  169a² – 144a²  = 169 x 169 – 144x 144

∴  25a²  = (169 + 144)(169 – 144)

∴  25a²  = (313)(25)

∴  a² = 313

∴  a = √313 = 17.69 m

Now T = 2π/ω = 2 x 3.14 /1 = 6. 28 s

Ans:  Period =6.28 s and amplitude = 17.69 cm

Example – 15:

A particle performing  S.H.M. has velocities of 8 cm/s and 6 cm/s at displacements of 3 cm and 4 cm respectively. Find its amplitude and frequency of oscillations. Calculate its maximum velocity. What is the phase of its motion when the displacement is 2.5 cm?

Solution:

Given: v₁ = 8 cm/s at x₁ =3 cm and v₂ = 6 cm/s at x₂ = 4 cm, displacement = x = 2.5 cm

To Find: Amplitude = a =? frequency = n = ?, phase = (ωt + α) =?,

∴  9a² – 81 = 16a² – 256

∴  16a² – 9a²  = 256 – 81

∴  7a²  = 175

∴  a²  = 25

∴  a = 5 cm

Now ω = 2 π n

∴ n = ω/2π = 2/( 2 x 3.142) = 0.3183 Hz

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  2.5 = 5 sin (ωt + α)

∴  sin (ωt + α) = 2.5/5 = 1/2

∴  (ωt + α) = sin^-1(1/2) = π/6

Ans: Amplitude is 5 cm, frequency = 0.3183 Hz, Phase = π/6 or 30°

Previous Topic: Numerical Problems on Displacement, Velocity, and Acceleration of Particle Performing S.H.M.

Next Topic: Graphical Representation of S.H.M.

Science > Physics > Oscillations: Simple Harmonic Motion > Numerical Problems on Maximum Velocity and Maximum Acceleration.

The post Numerical Problems on S.H.M. – 02 appeared first on The Fact Factor.

in this article, we shall study to calculate the initial phase, displacement, velocity, and acceleration of a body performing S.H.M.

Example – 1:

A particle performs a linear S.H.M along a path 10 cm long. The particle starts from a distance of 1 cm from the mean position towards the positive extremity. Find the epoch and the phase of motion when the displacement is 2.5 cm.

Given:  Path length = 10 cm, amplitude = path length /2 = 10/2 = 5 cm, Initial displacement = x₀ = 1 cm, Displacement = x = 2.5 cm.

To Find: Epoch = α =? and phase of S.H.M. = (ωt + α) = ?

Solution:

Epoch = α = sin^-1(x_o/a) = sin^-1(1/5) = sin^-1(0.2) = 11°32’

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  2.5 = 5 sin (ωt + α)

∴  sin (ωt + α) = 2.5/5 = 1/2

∴  (ωt + α) = sin^-1(1/2) = π/6

Ans: Initial phase is 11°32’ and phase of S.H.M. is π/6 or 30^o.

Example – 2:

The periodic time of a body executing S.H.M. is 2 s. After how much time interval from t =0 will its displacement be half the amplitude?

Given: Time period = T = 2 s, Displacement x = 1/2 a, particle passes through mean position, α = 0.

To Find: Time elapsed = t =?

Solution:

Angular velocity = ω = 2π/T = 2π/2 = π rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  1/2 a = a sin (πt + 0)

∴  1/2 =sin (πt)

∴  πt = sin^-1(1/2) = π/6

∴  t = 1/6 s

Ans: After 1/6 s the displacement will be half the amplitude

Example – 3:

A particle executes S.H.M.  of amplitude 25 cm and time period 1/3 seconds. What is the minimum time required for a particle to move between two points located at a distance of 12.5 cm on either side of the mean position?

Given: amplitude = a = 25 cm, x = 12.5 on either side, Period = T = 3 s, particle passes through mean position, α = 0.

To Find: Time required = 2t = ?

Solution:

Angular velocity = ω = 2π/T = 2π/3  rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  12.5 = 25 sin ((2π/3)t + 0)

∴  1/2 =sin ((2π/3)t)

∴  (2π/3)t = sin^-1(1/2) = π/6

∴  t = 1/4 s

Now the points are on either side of the mean position

Hence time taken to move between these two points = 2t = 2 x 1/4 = 0.5 s

Ans: the minimum time required is 0.5 s

Example – 4:

A particle executes S.H.M. of period 12 s and of amplitude 8 cm. What time will it take to travel 4 cm from the extreme position ?

Given: Period = T = 12 s, amplitude = a = 8cm, distance from extreme position = 4 cm, displacement = x = 8 cm – 4 cm = 4 cm, particle starts from extreme position, α = π/2.

To Find: time taken = t = ?, Velocity = v = ?

Solution:

Angular velocity = ω = 2π/T = 2π/12  = π/6 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  4 = 8 sin ((π/6)t + π/2)

∴  1/2 =cos ((π/6)t)

∴  (π/6)t = cos^-1(1/2) = π/3

∴  t = 2 s

Ans: Time taken = 2 s

Example – 5

A particle performs S.H.M. of period 4 s. If the amplitude of its oscillations is 4 cm, find the time it takes to describe 1 cm from the extreme position.

Given: Period = T = 4 s, amplitude = a = 4cm, distance from extreme position = 1 cm, displacement = x = 4 cm – 1 cm = 3 cm, particle starts from extreme position, α = π/2.

To Find: time taken = t = ?

Solution:

Angular velocity = ω = 2π/T = 2π/4  = π/2 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  3 = 4 sin ((π/2)t + π/2)

∴  3/4 = cos ((π/2)t)

∴  (π/2)t = cos^-1(3/4) = 41.41° = 41.41 x 0.0175= 0.7247 rad

∴  t = 0.7247 x 2 /3.142 = 0.4613 s

Ans:  Time taken = 0.4613 s

Example – 6

A particle performs S.H.M. of period 12 s along a path 16 cm long. If it is initially at the positive extremity, how much time will it take to cover a distance of 6 cm from that position?

Given: Period = T = 12 s, path length = 16 cm, amplitude = a = 16/2 = 8cm, distance from extreme position = 6 cm, displacement = x = 8 cm – 6 cm =2 cm, particle starts from extreme position, α = π/2.

To Find: time taken w.r.t. extreme position = t_e = ?

Solution:

Angular velocity = ω = 2π/T = 2π/12  = π/6 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  2 = 8 sin ((π/6)t + π/2)

∴  2/8 = cos ((π/6)t)

∴  (π/6)t = cos^-1(1/4) = 75.52° =75.52 x 0.0175 = 1.322 rad

∴  t = 1.322 x 6 /3.142 = 2.52 s

Ans: Time taken =2.52 s

Example – 7:

The shortest distance traveled by a particle performing S.H.M. from its mean position in 2 seconds is equal to √3/2  of its amplitude. Find its period.

Given: Time elapsed = t = 2s, displacement = x = a √3/2, particle passes through mean position, α = 0.

To Find: Period = T = ?

Solution:

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  a √3/2  = a sin (ωt + 0)

∴  √3/2  = sin ωt

∴  ωt = sin^-1(√3/2 ) = π/3

∴  (2π/T)t = π/3

∴  (2π/T)x 2 = π/3

∴  T = 2 x 2 x 3 = 12 s

Ans: Period is 12 s

Example – 8:

A particle performing S.H.M. has a period of 6 s and amplitude of 8 cm. The particle starts from the mean position and moves towards the positive extremity. Find its displacement, velocity, and acceleration 0.5 s after the start.

Given:  Period = T = 6 s, amplitude = a = 8cm, time elapsed = t = 0.5 s, particle starts from mean position, α = 0.

To Find: Displacement = x = ?, Velocity = v =?, acceleration = f = ?

Solution:

Angular velocity = ω = 2π/T = 2π/6 = π/3 rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  x = 8 sin ( π/3 x 0.5 + 0)

∴  x = 8 sin ( π/6) = 8 x 1/2 = 4 cm

The magnitude of the velocity of a particle performing S.H.M. is given by

The magnitude of the acceleration of a particle performing S.H.M. is given by

f = ω²x = (π/3)² x 4 =  (3.142/3)² x 4 = 4.38 cm/s².

Ans: Displacement = 4 cm; velocity = 7.26 cm/; acceleration = 4.38 cm/s²

Example – 9:

A sewing machine needle moves in a path 4 cm long and the frequency of its oscillations is 10 Hz. What are its displacement and acceleration 1/120 s after crossing the centre of its path?

Given: Path length = 4 cm, amplitude = path length/2 = 4/2 = 2 cm, Frequency of oscillation = n = 10 Hz, Time elapsed = t = 1/120 s,  particle passes through mean position, α = 0.

To Find: Displacement = x =? acceleration = f =?

Solution:

Agular velocity = ω = 2πn = 2π x 10 = 20π rad/s

Displacement of a particle performing S.H.M. is given by

x = a sin (ωt + α)

∴  x = 2 sin ( 20π x 1/120 + 0)

∴  x = 2 sin ( π/6) = 2 x 1/2 = 1 cm

The magnitude of the acceleration of a particle performing S.H.M. is given by

f = ω²x = (20π)² x 1 =  (20 x 3.142)² = 3944 cm/s².

Ans: Displacement =1 cm and acceleration = 3944 cm/s²

Previous Topic: Theory of Simple Harmonic Motion

Next Topic: Numerical Problems on Maximum Velocity and Maximum Acceleration of S.H.M.

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In this article, we shall study, the concept of linear simple harmonic motion (S.H.M.) and derive expressions for displacement, velocity, acceleration, and period of n object performing the linear simple harmonic motion.

Linear simple harmonic motion is defined as the motion of a body in which

• the body performs an oscillatory motion along its path.
• the force (or the acceleration) acting on the body is directed towards a fixed point (i.e. means position) at any instant.
• the force (or the acceleration) acting on a body is directly proportional to the displacement of the body at every instant.

e.g. the motion of heavy bob of the simple pendulum.

Examples of Linear Simple Harmonic Motion:

• The motion of the heavy bob of the pendulum.
• The motion of pivoted magnetic needle in the uniform magnetic field.
• The Motion of molecules of solid due to thermal energy.
• The motion of the needle of a sewing machine.
• The motion of prongs of tuning fork.
• The motion of medium particles, when a longitudinal or transverse wave travels through it.

Terminology of Linear Simple Harmonic Motion:

Displacement of Simple Harmonic Motion:

The distance of the body performing S.H.M. from its mean position is called as a displacement. It is denoted by the letter ‘x’. Its S.I. unit is metre.

Amplitude of Simple Harmonic Motion:

The maximum displacement of the body performing S.H.M. from the mean position is called as the amplitude of S.H.M. It is denoted by the letter ‘a’. Its S.I. unit is metre.

Path length or range of Simple Harmonic Motion:

The total length of the path over which the body performing linear S.H.M. moves is called the path length or range of linear S.H.M. Path length or range is twice the amplitude and in one oscillation body covers a total distance of ‘4a’. Thus, path length = 2 × amplitude

Period of Simple Harmonic Motion:

The time taken by the body performing S.H.M. to complete one oscillation is called as a period of S.H.M. It is denoted by a letter ‘T’ and its S.I. unit is second.

Frequency of Simple Harmonic Motion:

The number of oscillations performed by the body performing S.H.M in unit time (one second) is called a frequency of S.H.M. It is denoted by the letter ‘n’. Its S.I. unit is hertz (Hz).

Defining Equation of Linear Simple Harmonic Motion:

Linear simple harmonic motion is defined as the motion of a body in which

• the body performs an oscillatory motion along its path.
• the force (or the acceleration) acting on the body is directed towards a fixed point (i.e. means position) at any instant.
• the force (or the acceleration) acting on a body is directly proportional to the displacement of the body at every instant.

The force acting on the body, which is directed towards the mean position at every instant, it is called as a restoring force.

From the definition of S.H.M.

F  ∝  – x

∴ F     =   – k x

The negative sign indicated that it is restoring force,

Where,  F  =  restoring force, x  =  displacement

k is constant called force constant

This equation is known as the defining equation of linear S.H.M.

Differential Equation of Linear S.H.M.:

By definition of linear S.H.M., the force acting on a particle performing S.H.M. is given by

F = – kx     …..   (1)

Where k is force per unit displacement, which is constant.

The negative sign indicates that it is a restoring force.

By Newton’s second law of motion

F = m f       …..  (2)

Where m = mass of the body

f = acceleration of the body, F = restoring force acting on the body

By definition of linear acceleration,

From equations (2) and (5) we have

The equations (7) (8) and (9) (different forms) are known as differential equations of linear S.H.M. which is a second-order homogeneous differential equation.

Expression for Acceleration of a Particle Performing Linear S.H.M.:

The differential equation of S.H.M. is

Where k = Force constant, m = Mass of a body performing S.H.M.

This is an expression of an acceleration of a body performing linear S.H.M. Negative sign indicates the direction of acceleration towards the mean position or it is opposite to the direction of displacement.

Irrespective of the position of the particle performing S.H.M., the direction of acceleration is always towards the mean or equilibrium position.

Special Cases:

Case – I: When the body is at the mean position, its displacement x =  0.

f _min =  0

Thus at the mean position, acceleration is minimum and i.e. zero.

Case – II: When the body is at an extreme position. The magnitude of its displacement x = a

f =  ω²a

Thus at the extreme position, the magnitude of the acceleration is maximum

Expression for Velocity of a Particle Performing Linear S.H.M.:

The differential equation of S.H.M. is

Where k = Force constant, m = Mass of a body performing S.H.M.

Integrating both sides of the equation

Where C = Constant of integration, v = velocity of the body.

x = Displacement of the body.

At extreme position x = ± a   and v = 0

Substituting these values in equation (1) we have

This is an expression of the velocity of the particle performing S.H.M.

Where a = amplitude of S.H.M., x = displacement of body.

The direction of velocity may be towards the mean position or away from the mean position.

Special Cases:

Case – I: When particle is at mean position i.e. x = 0

The velocity of the particle performing S.H.M. is maximum at the mean position.

V_max =  ωa

Case – II: When particle is at extreme position i.e. x = a.

The velocity of the particle performing S.H.M. is minimum and i.e. zero at the extreme position

The expression for Displacement of a Particle Performing Linear S.H.M.:

The magnitude of the velocity of the particle performing S.H.M. is given by

where a  =  amplitude of S.H.M., x =  displacement of body.

ω =   Angular velocity

Integrating both sides

Case – 1: When the particle is starting from mean position, x = 0 at t = 0.

Substituting these values in equation (1)

This is an expression for displacement of the particle performing S.H.M. and starting from mean position.

Case 2: If particle is starting from extreme position i.e. x = ± a  at t = 0.

Substituting these values in equation (1)

This is is an expression for displacement of the particle performing S.H.M. and starting from an extreme position.

Case – 3: If the particle is starting from any other position than extreme or mean position. Let x = x₀  when t = 0.

Substituting these values in equation (1).

This is an expression for displacement of the particle performing S.H.M. and starting from any position.

Value of α depends upon the initial conditions. If body starts from mean position α =  0. If body starts from extreme position α =± π/2.

α is called an initial phase or epoch of S.H.M. and x₀ =  initial displacement of the particle. It is also denoted by letter ø_o.

The direction of displacement is always away from the mean position.

Expression for Velocity and Acceleration of a Particle Performing S.H.M. Using Expression of Displacement:

The general equation of displacement of a particle is

x = a sin (ωt + α)   ………….. (1)

The velocity of the particle can be obtained by differentiating both sides w.r.t. t,

This is an expression for the velocity of a particle performing linear S.H.M.

We can get the acceleration of the body by differentiating equation (2) again w.r.t. t.

This is an expression for the acceleration of a particle performing linear S.H.M.

Phase of S.H.M.:

Displacement of a particle performing S.H.M. is given by x = a sin (ωt + α)

where x  =  displacement, a  =  amplitude of S.H.M., =  angular velocity

t  =  time, α =  initial phase (epoch)

The quantity (ωt + α) is called as the phase of S.H.M.  Phase of a body performing linear S.H.M. is defined as the state of the body w.r.t. the mean position at instant ‘t’.

When the particle is at mean position quantity (ωt + α) is zero when it is at extreme position quantity (ωt + α) is  π/2. It means if particle starts moving from mean position to extreme position (ωt + α) starts increasing from zero to π/2.

When the body is at the mean position, then its phase is kπ where k = 0,1,2,3,….. and when body is at the extreme position then its phase is kπ.  where k  =  1,3,5,7,…..

From the value of (ωt + α), we can get an idea of the exact position and state of motion of the particle performing S.H.M.

Time Period of a Particle Performing Linear S.H.M.:

The differential equation of S.H.M. is

Where k = Force constant, m = Mass of a body performing S.H.M.

This is an expression for the time period of a particle performing linear S.H.M.

The expression for Time Period of a Particle Performing Linear S.H.M.  in Terms of Force Constant:

By differential equation of S.H.M.,

Where T = Period of S.H.M.

This is an expression for the time period of S.H.M. in terms of force constant.

Uniform circular motion is a special case of linear S.H.M.:

Consider particle ‘P’ moving along a circular path with uniform angular velocity w.  Let M be its projections on diameter AB of the circular path as shown in fig. The x components of displacement, velocity, and acceleration of particle P are same as x components of displacement, velocity, and acceleration of projection M respectively.

Suppose the particle P starts from the initial position with initial phaseα at time t  = 0. In time t the angle between OP and x-axis is (ωt + α).

From figure,

This is an expression for displacement of particle M at time t.

The velocity of the particle  M is given by

Thus the acceleration of particle M is directly proportional to the displacement of the particle and its direction is opposite to that of displacement. But this is the defining character of linear S.H.M. Hence uniform circular motion is a special case of linear S.H.M.

Restoring Force and Force Constant:

The force acting on the body performing linear S.H.M., which is directed towards the mean position (equilibrium position) at every instant, it is called as a restoring force.

From the definition of S.H.M.

F    ∝    – x

∴ F     =   – k x

The negative sign indicated that it is restoring force,

Considering magnitude only

F     =    k x

k is constant called force constant

∴  k = F/x = Restoring force/displacement

S.I. Unit of force constant is N/m.

Previous Topic: Introduction to Oscillations

Next Topic: Numerical Problems on Displacement, Velocity, and Acceleration of Particle Performing S.H.M.

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Angular Displacement:

For a particle performing a circular motion the angle, traced by the radius vector at the centre of the circular path in a given time is called the angular displacement of the particle at that time. It is denoted by ‘θ’. Its S.I. unit is radian (rad). It is a dimensionless quantity. [MºLºTº]

The direction of angular displacement: 

For smaller magnitude (infinitesimal) angular displacement is a vector quantity and its direction is given by the right-hand thumb rule.

Right-Hand Thumb Rule:

If we curl the fingers of our right hand and hold the axis of rotation with fingers pointing in the direction of motion then the outstretched thumb gives the direction of the angular displacement vector.

Sign Convention: 

An angular displacement in counter clock-wise direction is considered positive and that in the clockwise direction is considered as negative.

Vector relation between linear and angular displacement is

Characteristics of Angular Displacement:

• It is the angle, traced by the radius vector at the centre of the circular path in a given time is called the angular displacement of the particle at that time.
• For smaller magnitude (infinitesimal) angular displacement is a vector quantity and its direction is given by the right-hand thumb rule.
• Finite angular displacement is a vector quantity.
• Instantaneous angular velocity is a vector quantity.
• The direction of angular displacement in an anticlockwise sense is considered as positive, while the direction of angular displacement in a clockwise sense is considered as negative.
• The angular displacement of the particle performing uniform circular motion in equal time is equal.
• It is denoted by ‘θ’. Its S.I. unit is radian (rad). It is a dimensionless quantity. [MºLºTº].

Larger angular displacement is not treated as a vector quantity.

If a quantity has both the direction and magnitude then it seems to be vector quantity but it can only be treated as vector quantity if its satisfies laws of vector addition. Consider the following two cases of angular displacement

The commutative law of vector addition which states that if we add two vectors, the order in which we add them does not matter.

We can see that if the order is interchanged the final outcome is different. Thus the angular displacement fails to obey the law of vector addition. Hence larger angular displacement is not a vector quantity.

Angular Velocity:

The rate of change of angular displacement with respect to time is called the angular velocity of the particle. It is denoted by the letter ‘ω’. Its S.I. unit is radians per second (rad s^-1). Its dimensions are [MºLºT ^-1].

Mathematically,

For uniform circular motion, the magnitude of angular velocity is given by

Where ω = Angular speed, T = Period
N = Angular speed in r.p.m., n = Angular speed in r.ps. or Hz.
θ = Angular displacement, t = time taken

The Direction of Angular Velocity: 

For smaller magnitude (infinitesimal) the angular velocity is the vector quantity. Its direction is given by the right-hand thumb rule. It states that “If we curl the fingers of our right hand and hold the axis of rotation with fingers pointing in the direction of motion then the outstretched thumb gives the direction of the angular velocity vector”. Thus, the direction of angular velocity is the same as that of angular displacement.

By this rule, the direction of the angular velocity of the second hand, the minute hand, and the hour hand is perpendicular to the dial and directed inwards.

Angular Speed:

The angle traced by radius vector in unit time is called the angular speed or The magnitude of angular velocity is known an angular speed.

Uniform motion is that motion in which both the magnitude and direction of velocity remain constant. In UCM the magnitude of velocity is constant but its direction changes continuously. Hence UCM is not uniform motion. For uniform circular motion, the angular velocity is constant.

For uniform circular motion, the magnitude of velocity at P = magnitude of velocity at Q = magnitude of velocity at R and the direction of velocity at P ≠ direction of velocity at Q ≠ direction of velocity at R. In uniform circular motion a body moves in a circle describes equal angles in equal interval of time. Thus for a body performing UCM has uniform speed.

For non-uniform circular motion, The magnitude of velocity at P ≠ magnitude of velocity at Q ≠ magnitude of velocity at R and the direction of velocity at P ≠ direction of velocity at Q ≠ direction of velocity at R.  In non-uniform circular motion a body moves in a circle describes unequal angles in equal interval of time.

Characteristics of Angular Velocity:

• The rate of change of angular displacement with respect to time is called the angular velocity of the particle.
• Its direction is given by the right-hand thumb rule.
• The direction of angular velocity is the same as that of angular displacement.
• For uniform circular motion, the magnitude of angular velocity is constant.
• The magnitude of angular velocity (ω) is related to the magnitude of linear velocity (v) by the relation v = rω.
• It is denoted by the letter ‘ω’. Its S.I. unit is radians per second (rad s-1). Its dimensions are [MºLºT ^-1].

Example – 1:

The graph shows angular positions of a rotating disc at different instants. What is the sign of angular displacement and angular acceleration?

The angular velocity at any instant is given by ω = dθ/dt,

At t = 1 second the graph is rising up, thus the slope (dθ/dt) of the tangent at t = 1 second is positive. Hence angular velocity is positive.

At t = 2 seconds the graph reaches the topmost point, thus the slope (dθ/dt) of the tangent at t = 2 seconds is zero. Hence angular velocity is zero.

At t = 3 seconds the graph is going down, thus the slope (dθ/dt) of the tangent at t = 3 seconds is negative. Hence angular velocity is negative.

We can see the change in angular velocity as positive → zero → negative. Thus angular velocity is decreasing. Hence angular acceleration is negative.

Angular Acceleration: 

The average angular acceleration is defined as the time rate of change of angular velocity. It is denoted by the letter ‘α’. Its S.I. unit is radians per second square (rad /s²). Its dimensions are [MºLºT ^-2]. Mathematically,

If the initial angular velocity of the particle changes from initial angular velocity ω₁  to final ω₂ angular velocity in time ‘t’ then

The direction of Angular Acceleration:

The direction of angular acceleration is given by right-hand thumb rule. If the angular velocity is increasing then the angular acceleration has the same direction as that of the angular velocity.

If the angular velocity is decreasing then the angular acceleration has the opposite direction as that of the angular velocity.

For uniform circular motion angular acceleration is zero.

Characteristics of Angular Acceleration:

• The average angular acceleration is defined as the time rate of change of angular velocity.
• If the angular velocity is increasing angular acceleration is positive (e.g. the angular acceleration of the tip of a fan just switched on). If the angular velocity is decreasing angular acceleration is negative (e.g. the angular acceleration of the tip of a fan just switched off)
• If the angular velocity is increasing then the angular acceleration has the same direction as that of the angular velocity. If the angular velocity is decreasing then the angular acceleration has the opposite direction as that of the angular velocity.
• For uniform circular motion angular acceleration is zero.
• The magnitude of angular acceleration (α) is related to the magnitude of linear acceleration (a) by the relation a = rα.
• It is denoted by the letter ‘α’. Its S.I. unit is radians per second square (rad /s²). Its dimensions are [M⁰L⁰T^-2].

Right Handed Screw Rule:

When a right-handed screw is rotated in the sense of revolution of the particle, then the direction of the advance of the screw gives the direction of the angular displacement vector.

Relation Between Linear Velocity and Angular Velocity:

Consider a particle performing uniform circular motion, along the circumference of the circle of radius ‘r’ with constant linear velocity ‘v’ and constant angular speed ‘ω’ moving in the anticlockwise sense as shown in the figure.

Suppose the particle moves from point P to point Q through a distance ‘δx’along the circumference of the circular path and subtends the angle ‘δθ’ at the centre O of the circle in a small interval of time ‘δt’. By geometry

δx = r . δθ

If the time interval is very very small then arc PQ can be considered to be almost a straight line. Therefore the magnitude of linear velocity is given by

Thus the linear velocity of a particle performing uniform circular motion is radius times its angular velocity. In vector form above equation can be written as

The linear velocity can be expressed as the vector product of angular velocity and radius vector.

The following figure shows relative positions of the linear velocity vector, angular velocity vector, and radius or position vector.

For smaller magnitudes angular displacement, angular velocity are vector quantities. Let ( r) be the position vector of the particle at some instant. Let the angular displacement in small time δt be ( δθ). Let the corresponding linear displacement (arc length) be ( δs). By geometry

Dividing both sides of the equation by δt and taking the limit

Period of Revolution:

Let us consider particle performing a uniform circular motion. Let ‘T’ be its period of revolution. During the periodic time (T), particle covers a distance equal to the circumference 2pr of the circle with linear velocity v.

This is an expression for the period of revolution for particle performing the uniform circular motion.

The Expression for Angular Acceleration:

When a body is performing a non-uniform circular motion, its angular velocity changes. Hence the body possesses angular acceleration.
The rate of change of angular velocity w.r.t. time is called as the angular acceleration. We know that acceleration is the rate of change of velocity with respect to time.

r = radius of circular path = constant.

ω = angular velocity of the particle performing a circular motion

Where ‘α’ is angular acceleration. Hence,

linear acceleration = radius x angular acceleration.

If speed is increasing linear acceleration is in the same direction as that of linear velocity. If speed is decreasing linear acceleration is in the opposite direction to that of linear velocity. It is also referred as tangential acceleration. For uniform circular motion α = 0.

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