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 construction and working of sonometer, and its use to verify the laws of string.

Laws of Vibrating String:

Law of Length:

If the tension in the string and its mass per unit length of wire remains constant, then the frequency of transverse vibration of a stretched string is inversely proportional to the vibrating length.

Law of Tension:

If the vibrating length and mass per unit length of wire remain constant then, the frequency of transverse vibration of a stretched string is directly proportional to the square root of the tension in the string.

Law of Mass:

If the vibrating length and tension in the string remain constant then, the frequency of transverse vibration of a stretched string is inversely proportional to the square root of its mass per unit length

Sonometer:

Construction:

A sonometer consists of a hollow rectangular wooden box to which a uniform wire is attached at one end. The other end of the wire is passed over two horizontal knife edges or bridges and then over a pulley. A weight hanger is suspended from the free end of the wire. By placing different weights in the weight hanger, the tension in the wire can be suitably adjusted.

The points at which the wire rests on the knife edges cannot vibrate at all. Hence, when the wire is set up into vibrations, these two points become nodes and the wire vibrates in the fundamental mode. The frequency of vibration of the wire can be varied by either changing the positions of the knife edges by changing the vibrating length or by placing different weights in the pan by changing the tension.

Use of Sonometer to Determine the Unknown Frequency of a Tuning Fork:

To determine the unknown frequency of a tuning fork, the tension T in the wire is kept constant and the vibrating length between the knife edges is so adjusted, that the fundamental frequency of the wire becomes the same as that of the fork. To test this a small paper rider is placed on the wire midway between the knife edges where an antinode is formed.

The fork is set up into vibration and its stem is placed on the wooden box.  The length of the wire is adjusted till it vibrates in unison with the fork.  When this happens, the centre of the vibrating wire vibrates with maximum amplitude due to resonance, and the paper rider is thrown off. Then the frequency of the tuning fork which is the same as the fundamental frequency of the wire is given by

Where ‘m’ is the mass per unit length of the wire. The frequency of the fork is determined, knowing l, T, and m.

Use of Sonometer to Verify the Law of Length:

If the tension in the string and its mass per unit length of wire remains constant, then the frequency of transverse vibration of a stretched string is inversely proportional to the vibrating length.

To verify , the given wire (m = constant) is kept under constant tension (T = constant). A set of tuning forks having different frequencies n₁, n₂, n₃, n_4, etc. is taken. The length of the wire, vibrating in unison with each fork, is determined in turn using a paper rider or hearing beats. Let the lengths corresponding to the frequencies be l1, l2, l3, l4, etc.

Then, it is found that, within the limits of experimental error, n₁l₁ = n₂l₂ = n₃l₃ = n₄l₄ = constant. Thus in general n l = constant or n ∝ 1/l. If a graph of n  against  1/l  is plotted, it comes out as a straight line.

Use of Sonometer to Verify the Law of Tension:

If the vibrating length and mass per unit length of wire remain constant then, the frequency of transverse vibration of a stretched string is directly proportional to the square root of the tension in the string. To verify the law, the vibrating length of the given wire and linear density ‘m’ is constant.  A set of tuning forks having different frequencies n₁, n₂, n₃, n₄ etc. is taken.

By adjusting the tension T, each fork is made to vibrate in unison with the fixed length of the wire, one after the other. Let the tensions corresponding to the frequencies n1, n2, n3, n4, etc. be T1, T2, T3, T4, etc. respectively. Then it is found that, within limits of experimental error.

A graph of n² against T comes out as a straight line.

Use of Sonometer to Verify the Law of Mass:

If the vibrating length and tension in the string remain constant then, the frequency of transverse vibration of a stretched string is inversely proportional to the square root of its mass per unit length. This law cannot be verified directly, as either ‘n’ nor ‘m’ can be varied continuously as in the case of l or T. Therefore, this law is verified indirectly as follows.  The relation can be written as

Then, to verify the law, we must show that when n and T are kept constant.  A number of wires having linear densities m₁, m₂, m₃, m_4, etc. are taken.  Each one of them is subjected to the same tension T. Then, using a given tuning fork (n = constant) each wire is made to vibrate in unison with the fork, by adjusting its length. Let l1, l2, l3, l4, etc. be the vibrating length corresponding to linear densities m1, m2, m3, m4, etc. respectively.

Then it is found that, within the limits of experimental error.

Hence the law is indirectly verified.

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