JEE Physics 2023 Question with Solution

Given: The wave is $y(x, t) = 5 \sin(6t + 0.003x)$, where $x$ and $y$ are in cm and $t$ is in seconds.

Find: The wave velocity.

For a transverse harmonic wave, the speed is given by

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

Comparing $y(x, t) = 5 \sin(6t + 0.003x)$ with the general form $y(x,t)=A\sin(kx \pm \omega t)$, we identify:

• $A = 5 \, \text{cm}$
• $\omega = 6 \, \text{rad/s}$
• $k = 0.003 \, \text{rad/cm}$

Convert wave number into SI units:

$$k = 0.003 \, \text{rad/cm} = 0.3 \, \text{rad/m}$$

Now substitute the values:

$$v = \frac{6}{0.3} = 20 \, \text{m/s}$$

Therefore, the wave velocity is $20 \, \text{m/s}$.