Question

A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_1$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K_2$. Then

• A. $K_2=\frac{K_1}{4}$
• B. $K_2=2K_1$
• C. $K_2=K_1$
• D. $K_2=\frac{K_1}{2}$

Answer 1

The correct option is B $K_2=2K_1$

Calculate Kinetic energy $K_1$ and Kinetic energy $K_2$.

Maximum kinetic energy expression will be written as,

$\Rightarrow K_1=\frac{1}{2}m{\omega }^2{A}^2$

As we know,
Angular frequency $=\omega =\sqrt{\left(g/L\right)}$
Amplitude $A=L\theta$

$\Rightarrow K_1=\frac{1}{2}m{\left(\sqrt{\frac{g}{L}}\right)}^2(L\theta {)}^2$

$\Rightarrow K_1=\frac{1}{2}m\times \frac{g}{L}\times L^2{\theta }^2$

$\Rightarrow K_1=\frac{1}{2}mgL{\theta }^2\dots \left(i\right)$

If length is doubled,

$\Rightarrow L=2L$

$\Rightarrow K_2=\frac{1}{2}mg\left(2L\right){\theta }^2\dots \left(ii\right)$

Here, we are assuming the angular frequency is same for both cases
From both the equations,

$\Rightarrow \frac{K_1}{K_2}=\frac{\frac{1}{2}mgL{\theta }^2}{\frac{1}{2}mg\left(2L\right){\theta }^2}$

$\Rightarrow \frac{K_1}{K_2}=\frac{1}{2}$

$\Rightarrow K_2=2K_1$

Final answer: (c)