JEE Physics 2025 Question with Solution

Given: The wave equation is $y(x, t) = 4.0 \sin\left(20 \times 10^{-3} x + 600t\right)$ mm.

Find: The velocity of the wave.

Compare the given equation with the general form

$$y(x,t)=A\sin(kx+\omega t)$$

From this, we identify:

• Amplitude, $A = 4.0 \, \text{mm}$
• Wave number, $k = 20 \times 10^{-3} \, \text{rad/mm} = 20 \, \text{rad/m}$
• Angular frequency, $\omega = 600 \, \text{rad/s}$

The wave speed is

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

Substituting the values,

$$v = \frac{600}{20} = 30 \, \text{m/s}$$

This is the magnitude of the speed. Since the equation is of the form $\sin(kx + \omega t)$, the wave travels in the negative $x$-direction.

Therefore, the velocity of the wave is $-30 \, \text{m/s}$.

The correct option is C.

Given: The wave equation is $y(x, t) = 4.0 \sin\left(20 \times 10^{-3} x + 600t\right)$.

Find: The velocity of propagation.

For a travelling wave:

• $\sin(kx - \omega t)$ moves in the positive $x$-direction
• $\sin(kx + \omega t)$ moves in the negative $x$-direction

So the sign of the velocity is negative.

Now compute the magnitude:

$$|v| = \frac{\omega}{k} = \frac{600}{20} = 30 \, \text{m/s}$$

Hence,

$$v = -30 \, \text{m/s}$$

Therefore, the correct option is C.