Question

There is a pendulum 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 angular amplitude as in the first case, its maximum kinetic energy is $K_{2}$ . Then :

• A. $K_{2}$
• B. $K_{2}=2K_{1}$
• C. $K_{2}=K_{1}$
• D. $K_{2}=\frac{k_{1}}{4}$

Answer 1

Answer: (B)

Maximum kinetic energy $=\frac{1}{2}m\omega ^{2}A^{2}$
$\omega =\sqrt{\frac{g}{L}}$
$A=L\theta $
$KE=\frac{1}{2}m\frac{g}{L}\times L^{2}\theta ^{2},KE=\frac{1}{2}mgL\theta ^{2}$
$K_{1}=\frac{1}{2}mgL\theta ^{2}$
If Length Is Double
$K_{2}=\frac{1}{2}mg\left(\right.2L\left.\right)\left(\theta \right)^{2}$ [Here we are assuming angular amplitude is same]
$\frac{K_{1}}{K_{2}}=\frac{\frac{1}{2} m g l \left(\theta \right)^{2}}{\frac{1}{2} m g \left(\right. 2 L \left.\right) \left(\theta \right)^{2}}=\frac{1}{2}$
$K_{2}=2K_{1}$