In this article, we shall study techniques of measurement of time.

Time is the interval between two events. The study and science of time measurement is called chronometry or horology. Chronology, as opposed to chronometry, is the science of arranging events in their order or sequence of occurrence in time and is mainly used for studying the past (history).

The events which repeat after the same and equal interval of time are called periodic events. e.g. the motion of a bob of a simple pendulum, motion of needle of a sewing machine, etc.

In everyday practice, we use the 12-hour clock system. Airlines and Railways use the 24-hour clock system. 2:15 a.m. in 12-hour clock system equals 2:15 in the 24-hour clock system. 2:15 p.m. in a 12-hour clock system equals 14:15 in the 24-hour clock system.

Time can be measured using a simple pendulum, stopwatch, atomic clock. The time measuring instruments exhibit two basic components: (1) a regular, constant, or repetitive action to mark off equal increments of time, and (2) a means of keeping track of the increments of time and of displaying the result.

Time measurement is very important because the time can be considered as a common language between people and due to which everybody proceeds in an orderly manner in this fast-paced world. The time is important for keeping social interactions.

The oldest and most used method of measuring time is based on the earth rotating about its axis due to which the day and night occur. Rising and setting of the sun is considered as a periodic motion and is considered as the reference for measuring time. For primitive peoples, it was sufficient to divide the day into the early morning, mid-day, noon, late afternoon, night and late night. Due to further development of science and technology and due to more social interactions, there was a need to divide the day more precisely. The modern convention is to divide the day into 24 hours, an hour into 60 minutes, and a minute into 60 seconds (Babylonians 1900 b.c..–1650 b.c.). Babylonians have a great fantasy for number 12 and its multiples.

Old Time Measurement Techniques

Ghatika Yantra:

Ancient Indians devised a different type of clock, based on water, called as ghatika yantra. They divided day and night into 60 parts, each of which is called a ghari. The night and day are each divided into four parts each of which is called prahar. The kings used to have a group of men called ghariyalis appointed to measure time. A small vessel with a small hole at the bottom was placed over water contained in another big vessel. The water from the big vessel enters the small vessel with a hole. When the vessel with the hole was filled with water and sank, ghariyalis used to strike the ghariyal (the gong), a thick brass disc hung at a high place with a mallet. This indicated a certain period of time.

Sand Timers:

An hourglass (or sandglass, sand timer, or sand clock) is a device used to measure the duration of time. It is cannot be used to find the exact time but can be used to measure a predetermined time interval. It comprises two glass bulbs connected vertically by a narrow neck that allows a regulated trickle of sand from the upper bulb to the lower one.

Sand quantity, its coarseness, bulb size, and neck width determines the rate of flow of sand from the upper bulb to lower. They can be reused indefinitely by inverting the bulbs once the upper bulb is empty.

Sundial:

The simplest time measuring device in the old days was a stick put vertically in the sand. In the Greek language, the stick or the pillar whose shadow is used to measure the time is called the gnomon. The shadow of the stick was used to measure time.

Sundials are the oldest known devices that are used for time measurement. It depends on the rotation and movement of the sun. As the sun moves from east to west, the shadows formed change their position depending on the position of the sun at that instant using which we can predict the time of the day. The Egyptians were the first to use the sundials. Time was calculated depending on the length of the shadow. The plate on which the shadow of gnomon falls is called faceplate. This plate is flat but it can be shaped spherical, circular, conical or just about any shape. The face has markings on it to indicate the time.

Due to the tilted axis of the earth sundial will show a different time each week. The difference in time is compensated by aligning the gnomon with the earth’s axis.

The Jantar Mantar the largest stone sundial is constructed by Maharaja Jaisingh – II of Jaipur, India. It is an equinoctial sundial. Equinox is a day when the sun is exactly on the equator of the earth. The sundial consists of a gigantic triangular gnomon with its hypotenuse parallel to the Earth’s axis. On either side of the gnomon is a quadrant of a circle, parallel to the plane of the equator. It measures the time of day, corrects up to half a second.

Simple Pendulum:

Mechanized Clocks:

The escapement is a mechanism in a clock or watch that alternately checks and releases the train by a fixed amount and transmits a periodic impulse from the spring or weight to the balance wheel or pendulum. This mechanism was activated by falling weight or water.

The length of the pendulum is so adjusted that the cogwheel would move at the rate of one tooth per second.

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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 > Numerical Problems on a Simple Pendulum

Example – 01: 

The period of oscillation of a simple pendulum is 1.2 sec.in a place where g= 9.8m/s². How long is the bob below the rigid point?

Given: Period = T = 1.2 s, g = 9.8m/s².

To Find: Length of pendulum = l =?

Solution:

∴  l = 0.36 m

Ans: The bob is 0.36 m below the fixed point.

Example – 02:

If the period of oscillations of a simple pendulum is 4 s, find its length. If the velocity of the bob in the mean position is 40 cm/s, find its amplitude. g= 9.8 m/s².

Given: Period = T = 4 s, velocity at mean position = v_max = 40 cm/s, g = 9.8m/s².

To Find: Length of pendulum = l = ? amplitude =?

Solution:

∴ a = 25.5 cm

Ans: The length of the pendulum is 3.97 m and amplitude is 25.5 cm

Example – 03:

A simple pendulum of length 1 m has a mass of 10 g and oscillates freely with an amplitude of 2 cm. Find its PE at the extreme point. g = 9.8 m/s².

Given: Length of pendulum = l = 1 m, mass of bob = m = 10 g = 0.010 kg, amplitude = a = 2 cm = 0.02 m, g = 9.8m/s².

To Find: Potential energy at extreme point = E_P =?

Solution:

Ans: Potential energy at the extreme point is 1.96 x 10^-5 J

Example – 04:

Calculate the maximum velocity at which an oscillating pendulum of length one meter will attain if its amplitude is 8 cm. g = 9.8 m/s².

Given: length of pendulum = l = 1 m, amplitude = a = 8 cm.

To Find: Maximum velocity = v_max =?

Solution:

v_max = ωa = 3.13 x 8 = 25.04

Ans: maximum velocity is 25.04 cm/s

Example – 05:

When the length of a simple pendulum is decreased by 20 cm, the period changes by 10%. Find the original length and period of pendulum. g = 9.8 m/s²

Given: l ₂ m = l ₁ m – 20 cm = (l ₁ – 0.20) m,  g = 9.8m/s².

To Find: original length and period =?

Solution:

0.81 l = l – 0.20

l – 0.81 l = 0.20

0.19 l = 0.20

l = 0.20/0.19 = 1.05 m

Ans: Original length is 1.05 m and original period is 2.06 s.

Example – 06:

When the length of a simple pendulum is increased by 22 cm, the period changes by 20% .Find the original length of simple pendulum. g= 9.8 m/s²

Given: l ₂ m = l ₁ m + 22 cm = (l ₁ + 0.22) m,  g = 9.8 m/s². period change = 20%

To Find: original length and period = ?

Solution:

1.44 l = l + 0.22

1.44 l – l = 0.22

0.44 l = 0.22

l = 0.22/0.44 = 0.5 m

Ans: Original length is 0.5 m.

Example – 07:

The time of complete oscillation of a pendulum is doubled when the length of the pendulum is increased by 90 cm. Calculate the original length and original time of oscillation. g= 9.8 m/s².

Given: l ₂ m = l ₁ m + 90 cm = (l ₁ + 0.90) m,  g = 9.8m/s², T₂ = 2T₁.

To Find: original length and period = ?

Solution:

4 l = l + 0.90

4 l – l = 0.90

3 l = 0.90

l = 0.90/3 = 0.30 m= 30 cm

Ans: Original length is 30 cm. original period is 1.10 s

Example – 08:

The period of a simple pendulum at a place increases by 50%, when its length is increased by 0.6 m. Find its original length.

Given: l ₂ = l ₁ + 0.6 m,  g = 9.8 m/s².

To Find: original length = ?

Solution:

2.25 l = l + 0.6

2.25 l – l = 0.6

1.25 l = 0.6

l = 0.6/1.25 = 0.48 m

Ans: Original length is 0.48 m.

Example – 09:

The length of a seconds pendulum on the surface of the earth is one meter. What would be the length of a seconds pendulum on the surface of the moon where the acceleration due to gravity is g/6?

Given: Length on seconds pendulum on earth l_E = 1m, g_E = g anf g_M = g/6.

To Find: length of second’s pendulum on the Moon = ?

Solution:

Ans: Length of seconds pendulum on the Moon is 1/6 m

Example -10:

What should be the length of a simple pendulum on the surface of the moon if its period does not differ from that on the surface of the earth? Mass of earth is 80 times that of the moon and radius of the earth is 4 times that of the moon.

Given: M_M = 1/80 M_E, R_M = ¼ R_E, T_M = T_E = constant.

To Find: Length of the pendulum on moon =?

Solution:

Ans: The length of the pendulum on the moon is 1/5th of length pendulum on the earth.

Example – 11:

The mass and diameter of a planet are twice of the earth. What will be the period of oscillations of a pendulum on this planet if it is a seconds pendulum on earth?

Given: M_P = 2 M_E, R_P = 2 R_E, l_M = l_E = constant, T_E = 2 s.

To Find: Period of the pendulum on the planet = ?

Solution:

Ans: The period of the pendulum on the planet is 2.828 m.

Example – 12:

A pendulum takes 5 minutes for 100 oscillation at a place X. It takes the same time for 99 oscillations at another place Y. Compare the values of g at two places.

Given: Time period at place X = T_X = 5min/ 100 oscillations = 5 x 60/100 = 300/100 s, Time period at place Y = T_Y = 5min/ 99 oscillations = 5×60/99 = 300/99 s,

To Find: Ratio of g at two places g_X: g_Y=?

Solution:

Ans: The ratio of g at the two places is 1.02:1

Example – 13:

A clock regulated by a seconds pendulum keeps correct time. During summer the length of the pendulum increases to 1.01 m. How much will the clock gain or lose in one day? g= 9.8 m/s2.

Given: l₁ = 1m,  l₂ = 1.01 m , T₁ = 2 s.

To Find: Number of oscillations gain or lost =?

Solution:

Now the change in period = T₂ – T₁ = 2.017 – 2 = 0.017 s

For 2 seconds there is a decrease in 0.017 s

For 24 hours (24 x 60 x 60) the change is 0.017 x 24 x 60 x 60 /2 = 728.2 s

Ans:  A clock will Lose 728.2 s per day

Example – 14:

A clock pendulum of period one second has given a correct time when the length of the pendulum is 0.36 m. If the length of the pendulum is increased by 0.13 m, how much time will the clock gain or lose in 24 hours? g=9.8 m/s2 .

Given: l₁ = 0.36 m,  l₂ = 0.36 + 0.13 = 0.49 m ,

To Find: Number of oscillations gain or lost = ?

Solution:

Now the change in period = T₂ – T₁ = 1.405 – 1.204 = 0.201 s

For 1.204 seconds there is the increase of 0.201 s

For 24 hours (24 x 60 x 60) the change is 0.201 x 24 x 60 x 60 /1.204 = 14423 s = 4 hr

Ans:  A clock will lose 4 hours per day

Example – 15:

If the length of the second’s pendulum is decreased by 0.5% how many oscillations will it gain or lose in a day?

Given: l₂ = l₁ – 0.5 % l₁ = 0.995 l₁ , T₁ = 2 s.

To Find: Number of oscillations gain or lost =?

Solution:

T₂ = 0.9975 T₁ = 0.9975 x 2 = 1.995 s

Now the change in period = T1 – T2 = 2 – 1.995 = 0.005 s

For 2 seconds there is decrease of 0.005 s

For 24 hours (24 x 60 x 60) the change is 0.005 x 24 x 60 x 60 /2 = 216 s

Number of ocillations gained = 1/216 per second

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In this article, we shall study a concept of a simple pendulum, its characteristics. We shall also derive an expression for the period of a simple pendulum.

A simple pendulum consists of a point mass suspended from a perfectly rigid support by a weightless, inextensible, twist less and perfectly flexible fibre.  A heavy metal sphere having very very less radius than the length of the fibre can be treated as a point mass. This metal sphere is called as the bob of the simple pendulum. The distance between the point of suspension up to the centre of gravity of the bob is called as the length of the Simple pendulum.

When the simple pendulum is in the equilibrium position, the centre of gravity of bob lies vertically below the point of suspension. Hence position SA is the equilibrium position. Let the bob be displaced slightly (angular displacement less than 5° to one side and then released. The bob starts oscillating to and from about its equilibrium or the mean position. We can prove that this motion of the bob is linear S.H.M.

Let ‘l’ be the length of the simple pendulum and ‘m’ be the mass of bob of the simple pendulum. Let us consider bob at some arbitrary position B on its path. Let the displacement AB be equal to x. Let q be the corresponding angular displacement, i.e. m ∠ BOA = θ.

When the bob is at position B it is acted upon by following forces. The weight mg acting vertically downwards and the tension T’ in the string.  The weight mg can be resolved into two components. mg sinθ along the path of oscillation of bob and towards the mean position and mg cosθ along OB

As the string is inextensible, the component mg cos θ must be balanced by tension in the string.

T = mg Cosθ

The component mg sin θ is directed towards the mean or equilibrium position and remains unbalanced. Therefore it acts as restoring force.

F =  – mg Sinθ

The negative sign indicated that the force is opposite to the angular displacement

As angular displacement θ is very small

sin θ  ≈ θ   [θ measured in radian ]

F  =  –  m g θ    …  (1)

From figure and geometry of a circle

For particular place acceleration due to gravity ‘g’ is constant. Mass of bob of the pendulum is constant. As the string is inextensible, the length of the simple pendulum ‘’ is constant.  Hence quantity in the bracket is constant.

F ∝  -x

The above relation indicates that the force acting on the bob of the simple pendulum is directly proportional to the linear displacement which is defining a characteristic of simple harmonic motion. Hence the motion of the simple pendulum is linear S.H.M. for small amplitudes.

Period of Oscillation of Simple Pendulum:

Time taken by simple pendulum to complete one oscillation is called a Time period of the simple pendulum.

Force constant (k) is defined as restoring force per unit displacement.

This is an expression for the time period of oscillation of S.H.M.

Laws of a Simple Pendulum:

The period of a simple pendulum is given by

Where  l = Length of a simple pendulum, g = acceleration due to gravity

From the above expression, we can have the following conclusions.

• Law of Mass: As the expression doesn’t contain the term ‘m’, the time period of the simple pendulum is independent of the mass and material of the bob. This property is known as the law of mass.
• Law of Length: The time period of the simple pendulum is directly proportional to the square root of its length. This property is known as the law of length.
• Law of Iscochronism: The time period of the simple pendulum is independent of the amplitude, provided the amplitude is sufficiently small. This property is known as the law of isochronism. The oscillation of the simple pendulum is isochronous.
• Law of gravity: The time period of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity at that place. This property is known as the law of acceleration due to gravity.

The above conclusions are sometimes referred as laws of a simple pendulum.

The Frequency of Oscillation of a Simple Pendulum:

The period of a simple pendulum is given by

Where  l = Length of a simple pendulum, g = acceleration due to gravity

Now the frequency is related to period as n = 1 /T

Seconds Pendulum:

A simple pendulum whose time period is two seconds is called seconds pendulum. We know that

Squaring both sides we get

Thus the length of seconds pendulum is 0.99 m (approx. 1m)

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

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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.

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.

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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.

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

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

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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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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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