Oscillations

JEE Main level, single correct, medium

93 of 194

Question 93 of 194

A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1\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 K2\mathrm{K}_{2}

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Answer: (A) K2=2K1K_{2}=2 K_{1}

K=12mω2x2K=\frac{1}{2} m \omega^{2} x^{2}

⇒Kmax⁡=12mω2A2\Rightarrow K_{\max }=\frac{1}{2} m \omega^{2} A^{2}

A =Lθ=L \theta

ω=gL\omega=\sqrt{\frac{g}{L}}

⇒K=12m⋅gL⋅L2θ2\Rightarrow K=\frac{1}{2} m \cdot \frac{g}{L} \cdot L^{2} \theta^{2}

=12mgLθ2=\frac{1}{2} m g L \theta^{2}

∴K1K2=L2L=12⇒K2=2K1\therefore \frac{K_{1}}{K_{2}}=\frac{L}{2 L}=\frac{1}{2} \Rightarrow K_{2}=2 K_{1}

∴K1 K2=L2 L=12⇒K2=2 K1\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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