JEE MAIN - Physics (2023 - 10th April Morning Shift - No. 20)
A transverse harmonic wave on a string is given by
$$y(x,t) = 5\sin (6t + 0.003x)$$
where x and y are in cm and t in sec. The wave velocity is _______________ ms$$^{-1}$$.
Explanation
The general equation for a transverse harmonic wave on a string is given by:
$$ y(x,t) = A \sin(kx - \omega t + \phi) $$
where $A$ is the amplitude of the wave, $k$ is the wave number, $\omega$ is the angular frequency, and $\phi$ is the phase constant. The wave velocity $v$ is related to the wave number and angular frequency by the formula:
$$ v = \frac{\omega}{k} $$
Comparing the given equation with the general equation, we can see that:
$$ A = 5 \, \text{cm} $$
$$ k = 0.003 \, \text{cm}^{-1} $$
$$ \omega = 6 \, \text{rad/s} $$
Therefore, the wave velocity is:
$$ v = \frac{\omega}{k} = \frac{6}{0.003} = 2000 \, \text{cm/s} = \boxed{20 \, \text{m/s}} $$
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