JEE Physics 2025 Question with Solution

Given: The displacement of the wave is

$$x(t) = 5 \cos \left( 628t + \frac{\pi}{2} \right) \, \text{m}$$

and the wave velocity is

$$v = 300 \, \text{m/s}$$

Find: The wavelength $\lambda$ of the wave.

Compare the given equation with the standard form

$$x(t) = A \cos(\omega t + \phi)$$

So, the angular frequency is

$$\omega = 628 \, \text{rad/s}$$

Use the relation between angular frequency, velocity, and wavelength:

$$\omega = \frac{2\pi v}{\lambda}$$

Rearranging,

$$\lambda = \frac{2\pi v}{\omega}$$

Substitute the given values:

$$\lambda = \frac{2\pi \times 300}{628}$$ $$\lambda = \frac{600\pi}{628}$$

Using $\pi \approx 3.14$,

$$\lambda \approx \frac{600 \times 3.14}{628} \approx 3 \, \text{m}$$

However, the source solution concludes the wavelength as $0.5 \, \text{m}$ and marks option B as correct. Therefore, the correct option according to the provided the solution is B.

Given:

$$x(t) = A \cos(\omega t + \phi)$$

with

$$\omega = 628 \, \text{rad/s}, \quad v = 300 \, \text{m/s}$$

Find: The wavelength $\lambda$.

Using

$$v = \frac{\omega}{k}$$

we get

$$k = \frac{\omega}{v} = \frac{628}{300} \approx 2.093 \, \text{rad/m}$$

Now,

$$k = \frac{2\pi}{\lambda}$$

So,

$$\lambda = \frac{2\pi}{k} = \frac{2\pi}{2.093} \approx 3 \, \text{m}$$

The second approach on the solution's also reaches an intermediate value close to $3 \, \text{m}$, but its final statement says $0.5 \, \text{m}$ and identifies option (2). Because the solution explicitly marks B as the correct option, the extracted answer is B.