A simple harmonic oscillator has an amplitude A and time period 6 pi second. Assuming the oscillation starts from its mean position, the time required

Q: A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x}$ s, where $x=$__________.

a From phasor diagram particle has to move from $\mathrm{P}$ to $Q$ in a circle of radius equal to amplitude of $SHM.$ $ \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} $ $ \phi=\frac{\pi}{6}$ Now, $\frac{\pi}{6}=\omega t$ $\frac{\pi}{6}=\frac{2 \pi}{T} t$ $\frac{\pi}{6}=\frac{2 \pi}{6 \pi} t$ $\mathrm{t}=\frac{\pi}{2}$ So, $\mathrm{x}=2$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$2$
B$12$
C$4$
D$9$

JEE MAIN 2024, Diffcult

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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