Question:medium

A transverse wave is represented by $y = A \sin(kx - \omega t)$. The maximum particle velocity is

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Remember that for a sinusoidal wave, the maximum particle velocity is equal to the amplitude multiplied by the angular frequency.
Updated On: Jun 3, 2026
  • $A \omega$
  • $A k$
  • $\omega / k$
  • $A \omega k$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Understand the wave.
The wave is $y = A\sin(kx - \omega t)$. Here $y$ is how far a particle moves, $A$ is the amplitude, $k$ is the wave number, and $\omega$ is the angular frequency.

Step 2: Know particle velocity.
The particle velocity is how fast a point on the string moves up and down. We get it by differentiating $y$ with respect to time $t$, keeping $x$ fixed.

Step 3: Differentiate the wave.
\[ v_y = \frac{\partial y}{\partial t} = A\cos(kx-\omega t)\cdot(-\omega) \] So \[ v_y = -A\omega\cos(kx-\omega t) \]

Step 4: Find the maximum.
The cosine term swings between $-1$ and $+1$. The speed is largest when the cosine reaches its peak value of $1$.

Step 5: Write the maximum value.
Putting the cosine equal to one, \[ v_{y,max} = A\omega \]

Step 6: State the answer.
So the largest particle velocity is the amplitude times the angular frequency. \[ \boxed{v_{y,max} = A\omega} \]
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