How will the time period of simple pendulum change when the bob is replaced by another bob of?

A simple pendulum consists of a ball (point-mass) m hanging from a (massless) string of length L and fixed at a pivot point P. When displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. By applying Newton's secont law for rotational systems, the equation of motion for the pendulum may be obtained $$ \tau = I \alpha \qquad \Rightarrow \qquad -mg \sin\theta\; L = mL^2 \; \frac{d^2\theta}{dt^2} $$ and rearranged as $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0 $$ If the amplitude of angular displacement is small enough, so the small angle approximation ($\sin\theta\approx\theta$) holds true, then the equation of motion reduces to the equation of simple harmonic motion $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\theta = 0 $$ The simple harmonic solution is $$\theta(t) = \theta_o \cos(\omega t) \ , $$ where $\theta_o$ is the initial angular displacement, and $\omega = \sqrt{g/L}$ the natural frequency of the motion. The period of this sytem (time for one oscillation) is $$ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}} . $$

The period of a pendulum does not depend on the mass of the ball, but only on the length of the string. Two pendula with different masses but the same length will have the same period. Two pendula with different lengths will different periods; the pendulum with the longer string will have the longer period.

How many complete oscillations do the blue and brown pendula complete in the time for one complete oscillation of the longer (black) pendulum?

From this information and the definition of the period for a simple pendulum, what is the ratio of lengths for the three pendula?

With the assumption of small angles, the frequency and period of the pendulum are independent of the initial angular displacement amplitude. A pendulum will have the same period regardless of its initial angle. This simple approximation is illustrated in the animation at left. All three pendulums cycle through one complete oscillation in the same amount of time, regardless of the initial angle.

When the angular displacement amplitude of the pendulum is large enough that the small angle approximation no longer holds, then the equation of motion must remain in its nonlinear form $$ \frac{d^2\theta}{dt^2} + \frac{g}{L}\sin\theta = 0 $$ This differential equation does not have a closed form solution, but instead must be solved numerically using a computer. Mathematica numerically solves this differential equation very easily with the built in function NDSolve[ ].

The small angle approximation is valid for initial angular displacements of about 20° or less. If the initial angle is smaller than this amount, then the simple harmonic approximation is sufficient. But, if the angle is larger, then the differences between the small angle approximation and the exact solution quickly become apparent.

In the animation below left, the initial angle is small. The dark blue pendulum is the small angle approximation, and the light blue pendulum (initially hidden behind) is the exact solution. For a small initial angle, it takes a rather large number of oscillations before the difference between the small angle approximation (dark blue) and the exact solution (light blue) begin to noticeable diverge.

In the animation below right, the initial angle is large. The black pendulum is the small angle approximation, and the lighter gray pendulum (initially hidden behind) is the exact solution. For a large initial angle, the difference between the small angle approximation (black) and the exact solution (light gray) becomes apparent almost immediately.

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.

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