Question 167: Oscillations

Question

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 $$\frac{\pi}{x} \mathrm{~s}$$, where $$x=$$ _________.

Accepted Answer

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

Oscillations

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

Integer / Numeric Answer Type