Simple Pendulum: Derivation of the Time Period | SHM | Oscillation - Class 11 Physics

The Simple Pendulum is one of the simplest examples of Simple Harmonic Motion (SHM). It consists of a small heavy bob suspended from a fixed support by a light, flexible and inextensible string. When the bob is displaced slightly from its equilibrium position and released, it oscillates to and fro in a vertical plane. For small angular displacements, the motion of the pendulum is simple harmonic.

Definition of Simple Pendulum

A simple pendulum is an ideal mechanical system consisting of a point mass (called the bob) suspended by a light, massless and inextensible string from a rigid support. The bob is free to oscillate in a vertical plane under the action of gravity.

In practice, a true simple pendulum cannot be realized because the bob is not a perfect point mass and the string has a small mass. Therefore, a small metallic bob suspended by a light thread is considered a good approximation of a simple pendulum.

Construction of a Simple Pendulum

• A rigid support is fixed at the top.
• A light, flexible and inextensible string of length $L$ is suspended from the support.
• A small heavy bob of mass $m$ is attached to the lower end of the string.
• The bob is free to oscillate in a vertical plane.
• The lowest position of the bob is called the Mean Position.
• The maximum positions on either side are called the Extreme Positions.

Important Terms

• Length ($L$): Distance between the point of suspension and the centre of the bob.
• Angular Displacement ($\theta$): Angle made by the string with the vertical.
• Amplitude ($A$): Maximum displacement of the bob from its mean position.
• Time Period ($T$): Time taken by the bob to complete one full oscillation.
• Frequency ($f$): Number of oscillations completed in one second.

Motion of a Simple Pendulum

When the bob is displaced through a small angle $\theta$ and released, it oscillates about its mean position.

• At the extreme position, the velocity of the bob is zero.
• As it moves towards the mean position, its speed gradually increases.
• At the mean position, the velocity becomes maximum.
• After crossing the mean position, the speed decreases.
• Finally, the bob reaches the opposite extreme position and the motion repeats periodically.

For small angular displacements, this oscillatory motion is Simple Harmonic Motion (SHM).

Forces Acting on the Bob

When the bob is displaced through an angle $\theta$, only two forces act on it.

1. Tension ($T$) acting along the string towards the point of suspension.
2. Weight ($mg$) acting vertically downward.

The weight $mg$ can be resolved into two mutually perpendicular components.

1. Radial Component

The component of weight along the string is

$$mg\cos\theta$$

• Acts along the string.
• Opposes the tension.
• Together with tension, it provides the centripetal force.
• Produces no torque about the point of suspension.

The net radial force is

$$T-mg\cos\theta$$

2. Tangential Component

The component of weight perpendicular to the string is

$$mg\sin\theta$$

• Acts tangentially to the circular path.
• Always acts towards the mean position.
• Acts as the restoring force.

Hence, the restoring force is

$$F=-mg\sin\theta$$

The negative sign indicates that the restoring force is always opposite to the displacement.

Why is $mg\sin\theta$ called the Restoring Force?

The tangential component $mg\sin\theta$ always acts towards the equilibrium position regardless of the direction of displacement. It continuously tries to restore the bob to its mean position. Therefore, it is called the restoring force.

Proof that a Simple Pendulum Performs SHM (Force Method)

According to Newton's Second Law of Motion,

$$F=ma$$

The restoring force acting on the bob is

$$F=-mg\sin\theta$$

Therefore,

$$ma=-mg\sin\theta$$

or

$$a=-g\sin\theta$$

For small angular displacement,

$$\sin\theta\approx\theta$$

Hence,

$$a=-g\theta$$

From the geometry of the pendulum,

$$\theta=\frac{x}{L}$$

where

• $x$ = Linear displacement of the bob
• $L$ = Length of the pendulum

Substituting,

$$a=-g\left(\frac{x}{L}\right)$$

or

$$a=-\frac{g}{L}x$$

Since $\dfrac{g}{L}$ is constant,

$$a\propto-x$$

Thus, the acceleration of the bob is directly proportional to its displacement from the mean position and is always directed towards the mean position.

Hence, a simple pendulum executes Simple Harmonic Motion (SHM) for small angular displacements.

Key Points

• A simple pendulum performs SHM only for small angular displacements (approximately less than $20^\circ$).
• The restoring force is produced by the tangential component $$mg\sin\theta.$$
• The radial component $$mg\cos\theta$$ provides centripetal force but does not produce restoring torque.
• The restoring force is always directed towards the mean position.
• The negative sign in $$F=-mg\sin\theta$$ indicates that the restoring force always opposes the displacement.
• The equation $$a=-\frac{g}{L}x$$ is the mathematical condition for Simple Harmonic Motion.

Proof that a Simple Pendulum Performs SHM (Torque Method)

The force responsible for the oscillatory motion of the pendulum is the tangential component of its weight. Instead of using linear acceleration, we can also prove the SHM of a simple pendulum by applying the principle of rotational motion.

When the bob is displaced through a small angle $\theta$, the tangential component of weight is

$$mg\sin\theta$$

This component produces a restoring torque about the point of suspension.

The magnitude of the restoring torque is

$$\tau=-mgL\sin\theta$$

The negative sign indicates that the torque always acts opposite to the angular displacement and tends to restore the bob to its equilibrium position.

According to Newton's Second Law of Rotational Motion,

$$\tau=I\alpha$$

Substituting the value of torque,

$$I\alpha=-mgL\sin\theta$$

or

$$\alpha=-\frac{mgL}{I}\sin\theta$$

For small angular displacement,

$$\sin\theta\approx\theta$$

Therefore,

$$\alpha=-\frac{mgL}{I}\theta$$

Comparing this equation with the standard SHM equation,

$$\alpha=-\omega^2\theta$$

we obtain

$$\omega^2=\frac{mgL}{I}$$

Hence,

$$\boxed{\omega=\sqrt{\frac{mgL}{I}}}$$

Since angular acceleration is directly proportional to angular displacement and opposite in direction, the simple pendulum performs Simple Harmonic Motion (SHM).

Moment of Inertia of a Simple Pendulum

The bob of a simple pendulum is treated as a point mass situated at a distance $L$ from the point of suspension.

Therefore, the moment of inertia is

$$I=mL^2$$

Substituting this value into the angular frequency equation,

$$\omega=\sqrt{\frac{mgL}{mL^2}}$$

After simplification,

$$\boxed{\omega=\sqrt{\frac{g}{L}}}$$

Time Period of a Simple Pendulum

The relation between angular frequency and time period is

$$\omega=\frac{2\pi}{T}$$

Therefore,

$$T=\frac{2\pi}{\omega}$$

Substituting the value of angular frequency,

$$T=\frac{2\pi}{\sqrt{\frac{g}{L}}}$$

Hence,

$$\boxed{T=2\pi\sqrt{\frac{L}{g}}}$$

This is the time period of a simple pendulum for small oscillations.

Frequency of a Simple Pendulum

Frequency is the reciprocal of the time period.

$$f=\frac{1}{T}$$

Substituting the value of the time period,

$$\boxed{f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}}$$

Small Angle Approximation

The derivation of SHM for a simple pendulum is valid only when the angular displacement is very small.

The Taylor series expansion of $\sin\theta$ is

$$\sin\theta=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots$$

For very small values of $\theta$ (measured in radians), higher-order terms become negligible.

Hence,

$$\boxed{\sin\theta\approx\theta}$$

This approximation is generally valid for angular displacements less than about $20^\circ$.

Assumptions of a Simple Pendulum

• The string is light (massless).
• The string is perfectly inextensible.
• The bob is treated as a point mass.
• The oscillations are of small amplitude.
• Air resistance is neglected.
• The support is perfectly rigid.
• The motion takes place in a single vertical plane.
• The acceleration due to gravity remains constant throughout the motion.

Factors Affecting the Time Period

The time period depends upon the following quantities:

• Length of the pendulum ($L$)
• Acceleration due to gravity ($g$)

From the formula

$$T=2\pi\sqrt{\frac{L}{g}}$$

• If the length increases, the time period increases.
• If the acceleration due to gravity increases, the time period decreases.

Factors That Do Not Affect the Time Period

For small oscillations, the time period is independent of:

• Mass of the bob
• Material of the bob
• Shape and size of the bob
• Amplitude of oscillation (provided it is small)

Important Formulae

Restoring Force

$$F=-mg\sin\theta$$

Linear Acceleration

$$a=-\frac{g}{L}x$$

Restoring Torque

$$\tau=-mgL\sin\theta$$

Angular Acceleration

$$\alpha=-\omega^2\theta$$

Moment of Inertia

$$I=mL^2$$

Angular Frequency

$$\boxed{\omega=\sqrt{\frac{g}{L}}}$$

Time Period

$$\boxed{T=2\pi\sqrt{\frac{L}{g}}}$$

Frequency

$$\boxed{f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}}$$

Key Points

• The torque method is the standard NCERT derivation for proving SHM.
• The force method and torque method lead to the same result.
• The approximation $$\sin\theta\approx\theta$$ is essential for obtaining SHM.
• The time period of a simple pendulum is independent of the mass of the bob.
• The time period depends only on the length of the pendulum and the local value of acceleration due to gravity.
• The formula $$T=2\pi\sqrt{\frac{L}{g}}$$ is valid only for small oscillations.

Applications of a Simple Pendulum

• Used in pendulum clocks to measure time.
• Used to determine the local acceleration due to gravity.
• Used in seismometers to detect earthquakes.
• Used in metronomes for maintaining rhythm in music.
• Used in laboratories to study oscillatory motion and SHM.

Limitations of a Simple Pendulum

• A perfect simple pendulum cannot be constructed in practice.
• The formula for time period is valid only for small angular displacements.
• Air resistance and friction gradually reduce the amplitude.
• The string always possesses some mass.
• The bob cannot be treated as a perfect point mass.

Formula Summary

Quantity	Formula
Restoring Force	$$F=-mg\sin\theta$$
Linear Acceleration	$$a=-\frac{g}{L}x$$
Restoring Torque	$$\tau=-mgL\sin\theta$$
Moment of Inertia	$$I=mL^2$$
Angular Frequency	$$\omega=\sqrt{\frac{g}{L}}$$
Frequency	$$f=\frac{1}{2\pi}\sqrt{\frac{g}{L}}$$
Time Period	$$T=2\pi\sqrt{\frac{L}{g}}$$
Small Angle Approximation	$$\sin\theta\approx\theta$$

Multiple Choice Questions (MCQs)

1. The restoring force acting on a simple pendulum is
   • (A) $$mg$$
   • (B) $$mg\cos\theta$$
   • (C) $$mg\sin\theta$$
   • (D) $$T$$
2. The time period of a simple pendulum depends upon
   • (A) Mass of the bob
   • (B) Length of the pendulum and acceleration due to gravity
   • (C) Amplitude
   • (D) Shape of the bob
3. The SI unit of angular frequency is
   • (A) Hz
   • (B) rad s^-1
   • (C) m/s
   • (D) s
4. The approximation $$\sin\theta\approx\theta$$ is valid when
   • (A) $$\theta=90^\circ$$
   • (B) $$\theta$$ is very small
   • (C) $$\theta=180^\circ$$
   • (D) None of these
5. The motion of a simple pendulum is SHM because
   • (A) Restoring force is proportional to displacement
   • (B) Velocity is constant
   • (C) Acceleration is zero
   • (D) Force is constant

True or False

1. The restoring force acts towards the mean position. (True)
2. The time period depends upon the mass of the bob. (False)
3. The approximation $$\sin\theta\approx\theta$$ is valid only for small angles. (True)
4. A simple pendulum performs SHM for all amplitudes. (False)
5. The angular frequency is $$\sqrt{\frac{g}{L}}$$. (True)

Fill in the Blanks

1. The restoring force is __________ to the displacement. (Opposite)
2. The SI unit of frequency is __________. (Hertz)
3. The time period of a simple pendulum is $$__________. $$ \(2\pi\sqrt{\frac{L}{g}}\)
4. The restoring torque acts towards the __________ position. (Mean)
5. The approximation $$\sin\theta\approx\theta$$ is valid when the angle is measured in __________. (Radians)

Very Short Answer Questions

1. What is a simple pendulum?
2. Define amplitude.
3. What is the restoring force?
4. What is the SI unit of frequency?
5. State the small-angle approximation.

Short Answer Questions

1. Explain why a simple pendulum performs SHM.
2. Derive the expression for restoring force.
3. State the assumptions made in the theory of a simple pendulum.
4. What are the factors affecting the time period?
5. Why does the mass of the bob not affect the time period?

Long Answer Questions

1. Derive the expression for the time period of a simple pendulum using the torque method.
2. Prove that a simple pendulum performs Simple Harmonic Motion.
3. Derive the expression for angular frequency and frequency of a simple pendulum.

Numerical Problems

1. A simple pendulum has a length of $$1\,\text{m}$$. Calculate its time period if $$g=9.8\,\text{m s}^{-2}$$.
2. The time period of a pendulum is $$2\,\text{s}$$. Find its frequency.
3. Calculate the angular frequency of a pendulum of length $$0.49\,\text{m}$$.
4. A pendulum is taken from Earth to the Moon. How will its time period change?

Frequently Asked Questions (FAQs)

Q1. Why does a simple pendulum execute SHM?

Because for small angular displacement, the restoring force is directly proportional to the displacement and acts towards the mean position.

Q2. Why is the approximation $$\sin\theta\approx\theta$$ used?

It simplifies the restoring force equation into the standard SHM equation and is valid only for small angles measured in radians.

Q3. Does the mass of the bob affect the time period?

No. The time period depends only on the length of the pendulum and the acceleration due to gravity.

Q4. Why is the motion not SHM for large amplitudes?

For large angular displacements, $$\sin\theta$$ is no longer approximately equal to $$\theta$$. Therefore, the restoring force is not directly proportional to the displacement.

Q5. What happens to the time period if the length of the pendulum is doubled?

The time period becomes

$$T' = 2\pi\sqrt{\frac{2L}{g}}=\sqrt{2}\,T$$

Chapter Summary

• A simple pendulum is an ideal oscillating system.
• Its restoring force is $$F=-mg\sin\theta$$.
• For small angles, $$\sin\theta\approx\theta$$.
• Under this condition, the pendulum performs Simple Harmonic Motion.
• The angular frequency is $$\omega=\sqrt{\frac{g}{L}}$$.
• The time period is $$T=2\pi\sqrt{\frac{L}{g}}$$.
• The time period depends only on the length of the pendulum and the acceleration due to gravity.
• The formula is valid only for small oscillations.