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JEE MAINS 2024
31.01.24 S1
Question
A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is $\frac{2 \mathrm{~A}}{3}$. The new amplitude of motion is $\frac{\mathrm{nA}}{3}$. The value of n is
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Solution
Sol. $$ \begin{aligned} & \mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & \text { at } \mathrm{x}=\frac{2 \mathrm{~A}}{3} \end{aligned} $$ $$ v=\omega \sqrt{A^2-\left(\frac{2 A}{3}\right)^2}=\frac{\sqrt{5} A \omega}{3} $$ New amplitude $=\mathrm{A}^{\prime}$ $$ \begin{aligned} & \mathrm{v}^{\prime}=3 \mathrm{v}=\sqrt{5} \mathrm{~A} \omega=\omega \sqrt{\left(\mathrm{A}^{\prime}\right)^2-\left(\frac{2 \mathrm{~A}}{3}\right)^2} \\ & \mathrm{~A}^{\prime}=\frac{7 \mathrm{~A}}{3} \end{aligned} $$
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