QJEEM25Ph7cd7: JEE Main 2025 Physics - Class 11 JEEM25Ph7cd7 Solution: The equation of a transverse wave travelling along a string is…

Let's analyze the wave equation step by step.

The given wave is:

$y(x, t) = 4.0 \sin \left[20 \times 10^{-3} x + 600 t\right] \text{ mm}.$

First, simplify the coefficient of $x$ :

$20 \times 10^{-3} = 0.02 \, \text{mm}^{-1}.$

So the equation becomes:

$y(x, t) = 4.0 \sin\left(0.02x + 600t\right) \text{ mm}.$

A standard form for a travelling wave is:

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

which represents a wave moving in the positive $x$ -direction with speed $v = \frac{\omega}{k}$ .

Notice that our wave equation has the form:

$\sin(0.02x + 600t)$

The positive sign in front of $600t$ means we can rewrite the phase as:

$0.02x + 600t = 0.02x - (-600t),$

which indicates that the angular frequency $\omega$ in the standard form is effectively $-600$ .

The velocity $v$ of a wave is determined from the phase (for a constant phase, $\phi =$ constant):

$k x + \omega t = \text{constant}.$

Differentiating with respect to $t$ :

$k \frac{dx}{dt} + \omega = 0,$

which gives:

$\frac{dx}{dt} = -\frac{\omega}{k}.$

Substituting the values:

$k = 0.02 \, \text{mm}^{-1}$

$\omega = 600 \, \text{s}^{-1}$

We have:

$v = -\frac{600}{0.02} = -30000 \text{ mm/s}.$

Convert the velocity from mm/s to m/s:

$-30000 \, \text{mm/s} = -30 \, \text{m/s}.$

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

The correct answer is Option D.