Question

When a sound wave of wavelength $\lambda$ is propagating in a medium, the maximum velocity of the particle is equal to the wave velocity. The amplitude of wave is:

• A. $\lambda$
• B. $\frac{\lambda }{2}$
• C. $\frac{\lambda }{2\pi }$
• D. $\frac{\lambda }{4\pi }$

Answer 1

The correct option is C $\frac{\lambda }{2\pi }$

For all waves consisting of simple harmonic oscillation, the maximum velocity of particle is given as,
$\Rightarrow (v_P{)}_{\text{max}}=A\omega$
$A\to$amplitude of wave
Also the velocity of wave is given by,
$v=\frac{\omega }{k}=\frac{\omega }{\frac{2\pi }{\lambda }}$
$(\because k=\frac{2\pi }{\lambda })$
$\Rightarrow v=\frac{\lambda \omega }{2\pi }$
Given,
$(v_P{)}_{\text{max}}=v$
$\Rightarrow A\omega =\frac{\lambda \omega }{2\pi }$
$\Rightarrow A=\frac{\lambda }{2\pi }$