Question 29 of 194

A body is performing simple harmonic with an amplitude of $10cm$. The velocity of the body was tripled by air Jet when it is at $5cm$ from its mean position. The new amplitude of vibration is $\sqrt{x}cm$. The value of $x$ is _____ .

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Answer: 700

In SHM, velocity of the particle at any displacement, $x$ is given by, $v=\sqrt{{A}^{2}-{x}^{2}}$.
Therefore,
$\Rightarrow \dfrac{v}{3v}=\dfrac{\left(\sqrt{{A}_{1}^{2}-25}\right)}{\sqrt{{A}_{2}^{2}-25}}$ $\Rightarrow \sqrt{{A}_{2}^{2}-25}=3\times \sqrt{75}$
$\Rightarrow {A}_{2}^{2}=700$ $\Rightarrow {A}_{2}=\sqrt{700}cm$