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QJEEM24Ph86bd: JEE Main 2024 Physics - Class 11 JEEM24Ph86bd Solution: A simple harmonic oscillator has an amplitude and time period…

A simple harmonic oscillator has an amplitude AA and time period 6π6 \pi second. Assuming the oscillation starts from its mean position, the time required by it to travel from x=x= A to x=32x=\frac{\sqrt{3}}{2} A will be πx s\frac{\pi}{x} \mathrm{~s} , where x=x= _________.

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JEE Main 2024 (Online) 29th January Evening Shift Physics - Simple Harmonic Motion Question 15 English Explanation

From phasor diagram particle has to move from P\mathrm{P} to Q\mathrm{Q} in a circle of radius equal to amplitude of SHM.

amp;cos⁡ϕ=3 A2 A=32amp;ϕ=π6\begin{aligned}& \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} \\& \phi=\frac{\pi}{6}\end{aligned}

Now, π6=ωt\frac{\pi}{6}=\omega \mathrm{t}

amp;π6=2πTtamp;π6=2π6πt\begin{aligned}& \frac{\pi}{6}=\frac{2 \pi}{T} t \\& \frac{\pi}{6}=\frac{2 \pi}{6 \pi} t\end{aligned}

t=π2\mathrm{t}=\frac{\pi}{2}

So, x=2x=2

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