Question 18 of 194

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 $x=\dfrac{\sqrt{3}}{2}A$ will be $\dfrac{\pi }{x}s$, where x=_______.

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Answer: 2

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