Question

The displacement of a wave is represented by $y=0.6\times10^{-3}\sin(500t-0.05x)$ where all the quantities are in their proper units. The maximum particle velocity (in $\text{ms}^{-1}$) of the medium is

• A. 0.5
• B. 0.03
• C. 0.150
• D. 0.75
• E. 0.3

Hint:

Be careful not to confuse particle velocity ($A\omega$) with wave velocity ($\omega/k$). In this problem, wave velocity would be $500/0.05 = 10,000 \text{ ms}^{-1}$.

Answer 1

Answer: (E)

Concept:
For a traveling wave: \[ y = A \sin(\omega t - kx) \] [itemsep=6pt]
• Amplitude ($A$): Maximum displacement from mean position = $0.6 \times 10^{-3}$ m
• Angular Frequency ($\omega$): Rate of oscillation = 500 rad/s
• Particle Velocity ($v_p$): \[ v_p = \frac{\partial y}{\partial t} = A\omega \cos(\omega t - kx) \] Important Insight:
- The cosine function varies between $-1$ and $+1$ - Maximum velocity occurs when $\cos(\omega t - kx) = \pm 1$ Hence, \[ v_{\text{max}} = A\omega \]

Step 1: Substitute the given values
\[ v_{\text{max}} = (0.6 \times 10^{-3}) \times 500 \]

Step 2: Simplify step-by-step
\[ 0.6 \times 10^{-3} = 0.0006 \] \[ v_{\text{max}} = 0.0006 \times 500 \] \[ v_{\text{max}} = 0.3 \text{ m s}^{-1} \] Final Answer:
\[ \boxed{v_{\text{max}} = 0.3 \text{ m s}^{-1}} \]