Oscillations Question 5
Question 5 - 2024 (29 Jan Shift 2)
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 $\frac{\pi}{\mathrm{x}} \mathrm{s}$, where $\mathrm{x}=$
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Answer: (2)
Solution:
From phasor diagram particle has to move from $\mathrm{P}$ to $\mathrm{Q}$ 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, $\frac{\pi}{6}=\omega \mathrm{t}$
$\frac{\pi}{6}=\frac{2 \pi}{T} t$
$\frac{\pi}{6}=\frac{2 \pi}{6 \pi} t$
$\mathrm{t}=\frac{\pi}{2}$
So, $x=2$
