What will happen to the period of a simple pendulum if the bob is replaced by a bob with more mass?

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CONCEPT:

Simple pendulum:

• When a point mass is attached to an inextensible string and suspended from fixed support then it is called a simple pendulum.
• The time period of a simple pendulum is defined as the time taken by the pendulum to finish one complete oscillation.

\(⇒ T = 2\pi\sqrt{\frac{l}{g}}\)

• The above formula is only valid for small angular displacements.

Where, T = Time period of oscillation, l = length of the pendulum, and g = gravitational acceleration 

• Frequency of simple pendulum is the number of complete cycles in unit time.

\(\Rightarrow f=\frac{1}{T}\)

• The length of a simple pendulum is defined as the distance between the point of suspension to the center of the bob.

CALCULATION:

The frequency of the simple pendulum is given as,

\(\Rightarrow f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}\)     -----(1)

• By equation 1 it is clear that the frequency of a simple pendulum depends on the length and the gravitational acceleration, but it does not depend upon the mass of the bob.
• Therefore the frequency of the simple pendulum will remain the same when the mass of the bob of a pendulum is increased.
• Hence, option 3 is correct. 

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Let us take a general case where an object $O$ is oscillating from a fixed point $P$.

• $C$ is the center of mass of the object $O$.
• $P$ is the pivot point about which the object rotates (axis-of-rotation).

• $N$ is the normal force exerted on the object $O$ by the pivot $P$. $$$$

• Let the moment of inertia of the object $O$ be $I$.

• Let the distance between the point $P$ and point $C$ be $l$ (distance between the center of mass $C$ and the pivot $P$).

$$$$

We can write the torque equation for the object $O$ about the point $P$. The only forces which are acting on the object are the normal force ($N$) and the gravitational force ($mg$). Since the normal force $N$ is acting on the pivot, the torque due it is zero. Therefore, only gravitational force provides the torque.

$$\vec{\tau} = \vec{l} \times m\vec{g} \tag{1}$$

$$\tau = I\alpha = -mgl \sin\theta \tag{2}$$

The negative sign appears as the torque always tries to reduce the angle $\theta$.

Simplifying equation $(2)$ further, you get:

$$\frac{d^2 \theta}{dt^2} = -\frac{mgl}{I} \sin\theta \tag{3}$$

For small angles of $\theta$, the following approximation holds:

$$\sin \theta \approx \theta \tag{4}$$

Using $(4)$, you can rewrite $(3)$ as

$$\frac{d^2 \theta}{dt^2} = -\frac{mgl}{I} \theta \tag{5}$$

The general equation for a quantity $\phi$ varying harmonically without damping is given by:

$$\frac{d^2\phi}{dt^2} = - \omega^2\phi \tag{6}$$

If you observe carefully, the equation $(5)$ is similar to equation $(6)$. Comparing the two equations, you get:

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

$$\omega = \sqrt{\frac{mgl}{I}} \tag{7}$$

As you already know, the time period is related to $\omega$ as: $$T = \frac{2\pi}{\omega} \tag{8}$$

Substituting $\omega$ from $(7)$ in equation $(8)$, you get

$$T = 2\pi \sqrt{\frac{I}{mgl}} \tag{9}$$

The above equation gives the time period for any object oscillating harmonically without damping with small amplitudes.

A pendulum is just a special case of the above generalization.

Substiuting the values for the variables in equation $(9)$, you get:

$$T = 2\pi \sqrt{\frac{ml^2}{mgl}}$$

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

As you have seen in the derivation, $l$ is the distance between the pivot $P$ and the center of mass of the rotating object $O$. Therefore, moving the center of mass effectively changes the distance between the center of mass and the pivot. This causes a change in time period.