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

The post Composition of Two S.H.M.s appeared first on The Fact Factor.

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

Previous Topic: Energy of Particle Performing S.H.M.

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

The post Numerical Problems on Energy of Particle Performing S.H.M. appeared first on The Fact Factor.

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.

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

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

The post The Energy of Particle Performing S.H.M. appeared first on The Fact Factor.

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.

Previous Particle: Numerical Problems on Velocity and Acceleration of a Body Performing S.H.M.

Next Topic: Energy of Particle Performing S.H.M.

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.

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°

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