Question

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

• A. $A \omega$
• B. $A k$
• C. $\omega / k$
• D. $A \omega k$

Hint:

Remember that for a sinusoidal wave, the maximum particle velocity is equal to the amplitude multiplied by the angular frequency.

Answer 1

The Correct Option is A

Step 1: Concept
In a wave described by the equation \( y = A \sin(kx - \omega t) \), where \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency, the particle velocity can be found by differentiating the displacement with respect to time.

Step 2: Meaning
The maximum particle velocity in a wave occurs when the sine function reaches its peak derivative value of 1. This means we need to find the time-derivative of the given wave equation and then determine the maximum value it can take.

Step 3: Analysis
Given \( y = A \sin(kx - \omega t) \), let's differentiate this with respect to time \( t \): \[v_y = \frac{dy}{dt} = A \cos(kx - \omega t) \cdot (-\omega)\] This simplifies to: \[v_y = -A \omega \cos(kx - \omega t)\] The maximum value of the cosine function is 1. Therefore, the maximum particle velocity \( v_{y,\text{max}} \) occurs when \( \cos(kx - \omega t) = 1 \): \[v_{y,\text{max}} = A \omega\]

Step 4: Conclusion
The maximum particle velocity in the wave is given by \( A \omega \). Final Answer: (A)