Question

The equation of wave is $y = 60 \sin(1200t - 6x)$, where ' $y$ ' is in micron, ' $t$ ' is in second and ' $x$ ' is in metre. The ratio of maximum particle velocity to the wave velocity of wave propagation is

• A. 36
• B. $3.6 \times 10^{-5}$
• C. $3.6 \times 10^{-4}$
• D. $3.6 \times 10^{-6}$

Hint:

For a sinusoidal wave: \[ \frac{v_{\max}}{v}=Ak \] This avoids calculating the two velocities separately.

Answer 1

The Correct Option is B

Concept:
For wave: \[ y=A\sin(\omega t-kx) \] Maximum particle speed is: \[ v_{\max}=\omega A \] Wave speed is: \[ v=\frac{\omega}{k} \] So the required ratio is: \[ \frac{v_{\max}}{v}=\frac{\omega A}{\omega/k}=Ak \] ip

Step 1: Identify amplitude and wave number.
From \[ y=60\sin(1200t-6x) \] we get: \[ A=60\ \mu\text{m}=60\times10^{-6}\text{ m} \] and \[ k=6\text{ m}^{-1} \] ip

Step 2: Use the ratio formula.
\[ \frac{v_{\max}}{v}=Ak \] \[ =60\times10^{-6}\times 6 \] \[ =360\times10^{-6}=3.6\times10^{-4} \] So the direct calculation gives: \[ 3.6\times10^{-4} \] which matches option (C), not option (B). ip The direct calculation gives:
\[ \boxed{3.6\times10^{-4}} \]