JEE Physics 2026 Question with Solution

Given: The displacement is $x = A\sin\omega t$ and the time period is $T$.

Find: The value of $\beta$ if maximum potential energy occurs at $t = \frac{T}{2\beta}$.

Concept: In SHM, potential energy is

$$U = \frac{1}{2}kx^2$$

So potential energy is maximum when the magnitude of displacement is maximum, that is, when $|x| = A$.

From

$$x = A\sin\omega t$$

maximum potential energy occurs when

$$\sin\omega t = \pm 1$$

The first time for this is

$$\omega t = \frac{\pi}{2}$$

Therefore,

$$t = \frac{\pi}{2\omega}$$

Using

$$\omega = \frac{2\pi}{T}$$

we get

$$t = \frac{\pi}{2}\cdot\frac{T}{2\pi} = \frac{T}{4}$$

Given that

$$t = \frac{T}{2\beta}$$

So,

$$\frac{T}{2\beta} = \frac{T}{4}$$

Hence,

$$2\beta = 4$$

which gives

$$\beta = 2$$

Therefore, the value of $\beta$ is $2$.

Given: The oscillator follows $x = A\sin\omega t$.

Find: The value of $\beta$.

Potential energy in SHM depends on displacement squared, so it is greatest at the extreme positions. Thus the required condition is that the particle must be at $x = A$ or $x = -A$.

From

$$x = A\sin\omega t$$

we require

$$A\sin\omega t = \pm A$$

so

$$\sin\omega t = \pm 1$$

The earliest positive time occurs for

$$\omega t = \frac{\pi}{2}$$

Thus,

$$t = \frac{\pi}{2\omega}$$

Now substitute

$$\omega = \frac{2\pi}{T}$$

Then

$$t = \frac{\pi}{2\left(\frac{2\pi}{T}\right)}$$

which simplifies to

$$t = \frac{\pi T}{4\pi} = \frac{T}{4}$$

The question gives

$$t = \frac{T}{2\beta}$$

Equating the two expressions:

$$\frac{T}{2\beta} = \frac{T}{4}$$

Cancel $T$ from both sides:

$$\frac{1}{2\beta} = \frac{1}{4}$$

Therefore,

$$2\beta = 4$$

and hence

$$\beta = 2$$

Therefore, the final answer is $2$.