Question

A transverse wave is represented by the equation $y=y_0 \sin\frac{2\pi}{\lambda}(vt-x)$ For what value of $\lambda$, is the maximum particle velocity equal to two times the wave velocity?

• A. $\lambda=\frac{\pi y_0}{2}$
• B. $\lambda=\frac{\pi y_0}{3}$
• C. $\lambda=2\pi y_0$
• D. $\lambda=\pi y_0$

Answer 1

The Correct Option is D

To solve this problem, we begin by understanding the given wave equation:

y = y_0 \sin \frac{2\pi}{\lambda}(vt - x)

Here, y_0 is the amplitude of the wave, \lambda is the wavelength, v is the wave velocity, and x is the position.

The particle velocity v_p is given by the derivative of y with respect to time t:

v_p = \frac{\partial y}{\partial t} = \frac{\partial}{\partial t} \left( y_0 \sin \frac{2\pi}{\lambda}(vt - x) \right)

Using the chain rule for differentiation, this becomes:

v_p = y_0 \left( \frac{2\pi}{\lambda} \right) v \cos \frac{2\pi}{\lambda}(vt - x)

For maximum particle velocity, the maximum value of \cos function is 1. Thus,

v_{\text{max}} = y_0 \frac{2\pi v}{\lambda}

The problem states that this maximum particle velocity is two times the wave velocity, v. Therefore,

v_{\text{max}} = 2v

Substituting the expression for v_{\text{max}}, we get:

y_0 \frac{2\pi v}{\lambda} = 2v

Canceling v from both sides (since v ≠ 0), we obtain:

y_0 \frac{2\pi}{\lambda} = 2

Solving for \lambda, we find:

\lambda = \frac{2\pi y_0}{2} = \pi y_0

Thus, the correct value of \lambda is \pi y_0, matching with the given correct answer.