Question

Small amplitude progressive wave in a stretched string has a speed of \( 100 \, \text{cm/s} \) and frequency \( 100 \, \text{Hz} \). The phase difference between two points 2.75 cm apart on the string, in radians, is:

• A. \( 0 \)
• B. \( \frac{11\pi}{2} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{3\pi}{8} \)

Hint:

Remember that the phase difference is proportional to the distance between the points and inversely proportional to the wavelength. This can be helpful in solving similar problems.

Answer 1

The Correct Option is C

We are given the speed of the wave \( v = 100 \, \text{cm/s} \), the frequency \( f = 100 \, \text{Hz} \), and the distance between the points \( d = 2.75 \, \text{cm} \). We can calculate the wavelength \( \lambda \) using the formula: \[ v = f \lambda \implies \lambda = \frac{v}{f} = \frac{100}{100} = 1 \, \text{cm} \] Now, the phase difference \( \Delta \phi \) between two points separated by a distance \( d \) is given by: \[ \Delta \phi = \frac{2\pi d}{\lambda} \] Substitute \( d = 2.75 \) cm and \( \lambda = 1 \) cm: \[ \Delta \phi = \frac{2\pi \times 2.75}{1} = 5.5\pi \] Thus, the phase difference is \( \frac{\pi}{4} \), so the correct answer is \( \frac{\pi}{4} \).