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

Correct answer: A. K_ 2 =2 K_ 1 K=(1)/(2) m ω^ 2 x^ 2 K_ =(1)/(2) m ω^ 2 A^ 2 A =L θ ω= (g)/(L) K=(1)/(2) m · (g)/(L) · L^ 2 θ^ 2 =(1)/(2) m g L θ^ 2 K_ 1 K_ 2 =(L)/(2 L)=(1)/(2) K_ 2 =2 K_ 1 K_ 1 ~K_ 2 = L 2 ~L =(1)/(2) K_ 2 =2 ~K_ 1

JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Physics
Chapter: Oscillations

JEE Main 2019 · 11 January · Evening Shift · Physics · Oscillations

A pendulum is executing simple harmonic motion and its maximum...

A pendulum is executing simple harmonic motion and its maximum kinetic energy is

\mathrm{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

\mathrm{K}_{2}

Options

A $K_{2}=2 K_{1}$
B $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{2}$
C $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{4}$
D $K_{2}=K$

Correct answer

A. $K_{2}=2 K_{1}$

Step-by-step solution

K=\frac{1}{2} m \omega^{2} x^{2}

\Rightarrow K_{\max }=\frac{1}{2} m \omega^{2} A^{2}

=L \theta

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

\Rightarrow K=\frac{1}{2} m \cdot \frac{g}{L} \cdot L^{2} \theta^{2}

=\frac{1}{2} m g L \theta^{2}

\therefore \frac{K_{1}}{K_{2}}=\frac{L}{2 L}=\frac{1}{2} \Rightarrow K_{2}=2 K_{1}

\therefore \frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}=\frac{\mathrm{L}}{2 \mathrm{~L}}=\frac{1}{2} \Rightarrow \mathrm{K}_{2}=2 \mathrm{~K}_{1}

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