JEE Physics 2024 Question with Solution

Given: Amplitude is $A$, time period is $T = 6\pi \, \text{s}$, and the oscillator starts from the mean position.

Find: The value of $x$ if the time taken to travel from $x = A$ to $x = \frac{\sqrt{3}}{2}A$ is $\frac{\pi}{x}$ seconds.

For SHM starting from the mean position, we can use the reference-circle relation. When the particle moves from the extreme position $x = A$ to $x = \frac{\sqrt{3}}{2}A$, we write

so

Hence the corresponding time interval satisfies

Using

we get

Therefore,

Since $T = 6\pi$ seconds,

Now compare with the given form

So,

Thus,

Therefore, the correct option is C.

Given: $T = 6\pi \, \text{s}$ and displacement changes from $A$ to $\frac{\sqrt{3}}{2}A$.

Find: The value of $x$ in $t = \frac{\pi}{x}$.

At the extreme position, the SHM phase for cosine representation is $0$. Moving to displacement $\frac{\sqrt{3}}{2}A$ means

This gives

The time for a phase change $\theta$ is

Now,

So,

Thus,

which gives

Therefore, the correct option is C. The solution also concludes $x = 2$.