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QJEEM24Ph5cd7: JEE Main 2024 Physics - Class 11 JEEM24Ph5cd7 Solution: A particle executes simple harmonic motion with an amplitude of …

A particle executes simple harmonic motion with an amplitude of 4 cm4 \mathrm{~cm} . At the mean position, velocity of the particle is 10 cm/s10 \mathrm{~cm} / \mathrm{s} . The distance of the particle from the mean position when its speed becomes 5 cm/s5 \mathrm{~cm} / \mathrm{s} is α cm\sqrt{\alpha} \mathrm{~cm} , where α=\alpha= ________.

Step-by-Step Verified Solution

amp;Vat mean position =Aω⇒10=4ωamp;ω=52amp; V=ωA2−x2amp;5=5242−x2⇒x2=16−4amp;x=12 cm\begin{aligned}& \mathrm{V}_{\text {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \\& \quad \omega=\frac{5}{2} \\& \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\& 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \\& \mathrm{x}=\sqrt{12} \mathrm{~cm}\end{aligned}

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