Question

The equation of a simple harmonic wave is given by $y=3 \sin \frac{\pi}{2}(50t-x), $ where $x$ and $y$ are in metres and $t$ is in seconds. The ratio of maximum particle velocity to the wave velocity is

• A. $2\pi$
• B. $\frac{3}{2}\pi$
• C. $3\pi$
• D. $\frac{2}{3}\pi$

Answer 1

The Correct Option is B

We are given the equation of a simple harmonic wave as:

y = 3 \sin \left( \frac{\pi}{2}(50t-x) \right)

Where:

• x and y are in metres
• t is in seconds

We need to find the ratio of the maximum particle velocity to the wave velocity.

The general form of a wave equation is:

y = A \sin (k x - \omega t)

Where:

• A is the amplitude of the wave
• k is the wave number
• \omega is the angular frequency

Comparing the given equation with the standard form:

\omega = \frac{\pi}{2} \times 50 = 25\pi

k = \frac{\pi}{2}

The wave speed v is given by:

v = \frac{\omega}{k} = \frac{25\pi}{\pi/2} = 50 \, \text{m/s}

The maximum particle velocity v_{\text{max}} is the product of the amplitude and the angular frequency:

v_{\text{max}} = A \omega = 3 \times 25\pi = 75\pi \, \text{m/s}

Therefore, the ratio of the maximum particle velocity to the wave velocity is:

\text{Ratio} = \frac{v_{\text{max}}}{v} = \frac{75\pi}{50} = \frac{3}{2}\pi

Thus, the correct answer is:

\frac{3}{2}\pi