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PhysicsSHMMiscellaneousMedium2 minPYQ_2022

PhysicsMediumnumerical

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

Answer:

700.00

Solution:

In SHM, velocity of the particle at any displacement, $x$ is given by, $v = \sqrt{A^{2} - x^{2}}$ .

Therefore,

$\Rightarrow \frac{v}{3 v} = \frac{(\sqrt{A_{1}^{2} - 25})}{\sqrt{A_{2}^{2} - 25}}$ $\Rightarrow \sqrt{A_{2}^{2} - 25} = 3 \times \sqrt{75}$

$\Rightarrow A_{2}^{2} = 700$ $\Rightarrow A_{2} = \sqrt{700} cm$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2022

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