Question

A transverse wave is described by the equation \( y = A \sin \left( 2 \pi \left( \frac{r}{\lambda} - \frac{t}{T} \right) \right) \). The maximum particle velocity is equal to four times the wave velocity if:

• A. \( \lambda = 2A \)
• B. \( \lambda = \pi A \)
• C. \( \lambda = \frac{\pi A}{2} \)
• D. \( \lambda = \frac{\pi A}{4} \)

Hint:

To relate particle velocity and wave velocity, remember the basic relationships for each, and then solve the equation based on the given conditions.

Answer 1

The Correct Option is C

The wave equation is given as \( y = A \sin \left( 2 \pi \left( \frac{r}{\lambda} - \frac{t}{T} \right) \right) \), where: - \( A \) is the amplitude, - \( \lambda \) is the wavelength, - \( T \) is the period, and - \( r \) is the position. The maximum particle velocity is given by \( v_p = \omega A \), where \( \omega \) is the angular frequency. The wave velocity is \( v_w = \frac{\omega}{k} \), where \( k = \frac{2\pi}{\lambda} \) is the wave number. Given that the maximum particle velocity is equal to four times the wave velocity, we have the equation: \[ v_p = 4 v_w \] Substituting \( v_p = \omega A \) and \( v_w = \frac{\omega}{k} \), we get: \[ \omega A = 4 \times \frac{\omega}{k} \] Simplifying, we find: \[ A = 4 \times \frac{1}{k} \] Since \( k = \frac{2\pi}{\lambda} \), substituting this value: \[ A = 4 \times \frac{\lambda}{2\pi} \] Solving for \( \lambda \), we get: \[ \lambda = \frac{\pi A}{2} \] Thus, the correct answer is \( \lambda = \frac{\pi A}{2} \).