JEE Physics 2025 Question with Solution

Given: $x(t) = x_0 \sin^2 \left( \frac{\pi t}{T} \right)$ with $x_0 = 1 \, \text{m}$.

Find: The correct graph of kinetic energy $K$ as a function of displacement $x$.

The solution states that the correct option is A.

For the motion,

$$v(t) = \frac{d}{dt}\left(x_0 \sin^2 \left( \frac{\pi t}{T} \right)\right)$$

Using differentiation,

$$v(t) = 2x_0 \sin \left( \frac{\pi t}{T} \right) \cos \left( \frac{\pi t}{T} \right) \frac{\pi}{T}$$

Now use $2\sin\theta\cos\theta = \sin 2\theta$,

$$v(t) = \frac{\pi x_0}{T} \sin \left( \frac{2\pi t}{T} \right)$$

Hence,

$$K = \frac{1}{2}mv^2$$

so the kinetic energy is zero when the velocity is zero, and maximum at the मध्य position of the motion. Since $x = x_0 \sin^2 \left( \frac{\pi t}{T} \right)$ varies from $0$ to $x_0 = 1$, the graph of $K$ versus $x$ starts from zero at $x=0$, becomes maximum at $x = \frac{1}{2}$, and returns to zero at $x=1$.

Therefore, the correct option is A (Graph 1).

Given: $x(t) = x_0 \sin^2 \left( \frac{\pi t}{T} \right)$.

Find: The shape of the graph of $K$ as a function of $x$.

Let

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

Then,

$$x = x_0 \sin^2 \theta$$

and from differentiation,

$$v = 2x_0 \sin \theta \cos \theta \frac{\pi}{T}$$

So,

$$v^2 = 4x_0^2 \left(\frac{\pi}{T}\right)^2 \sin^2 \theta \cos^2 \theta$$

Using

$$\sin^2 \theta = \frac{x}{x_0}, \qquad \cos^2 \theta = 1 - \frac{x}{x_0}$$

we get

$$v^2 \propto \frac{x}{x_0}\left(1 - \frac{x}{x_0}\right)$$

Therefore,

$$K \propto x\left(1 - \frac{x}{x_0}\right)$$

With $x_0 = 1$,

$$K \propto x(1-x)$$

This is a downward-opening parabola with zeros at $x=0$ and $x=1$, and maximum at

$$x = \frac{1}{2}$$

So the required graph is the arch-shaped curve shown in Graph 1.

Therefore, the correct option is A.