JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 26)
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 $$x=$$ _________.
Answer
2
Explanation
_29th_January_Evening_Shift_en_26_1.png)
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, $$x=2$$
Comments (0)
