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

Answer 1

Answer: (C)

$K_2=2K_1$

Maximum kinetic energy $=1/2m{\omega }^2{A}^2$

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

$A=L\theta$

$KE=1/2m\frac{g}{L}\times L^2{\theta }^2$, KE $=1/2mgL{\theta }^2$

$K_1=1/2mgL{\theta }^2$

If length is doubled

$K_2=1/2mg\left(2L\right){\theta }^2$

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

$K_2=2K_1$