JEE PYQ: Oscillations Question 41
Question 41 - 2019 (11 Jan 2019 Shift 2)
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 $K_2$.
(1) $K_2 = 2K_1$
(2) $K_2 = \frac{K_1}{2}$
(3) $K_2 = \frac{K_1}{4}$
(4) $K_2 = K_1$
Show Answer
Answer: (1)
Solution
$K_{max} = \frac{1}{2}m\frac{g}{L}A^2$. Doubling $L$: $\frac{K_1}{K_2} = \frac{L}{2L} = \frac{1}{2}$. $K_2 = 2K_1$.
BETA
