Question

In a plane progressive harmonic wave the maximum particle speed is always less than the wave speed if:

• A. Amplitude of wave $<\frac{\lambda }{2\pi }$
• B. Amplitude of wave $>\frac{\lambda }{2\pi }$
• C. Amplitude of wave $<\lambda$
• D. Amplitude of wave $>\frac{\lambda }{\pi }$

Answer 1

The correct option is A Amplitude of wave $<\frac{\lambda }{2\pi }$

Let the equation of a plane progressive harmonic wave be:
$y=\text{A sin}(\omega t-kx+\varphi )$
Wave speed is, $v=\frac{\omega }{k}$
and maximum particle speed is given by,
$(v_P{)}_{\text{max}}=A\omega$
For maximum particle speed to be always less than the wave speed,
$\Rightarrow (v_P{)}_{\text{max}}<v$
$\Rightarrow A\omega <\frac{\omega }{k}$
or $A<\frac{1}{k}$
$\because k=\frac{2\pi }{\lambda }$
$\Rightarrow A<\frac{\lambda }{2\pi }$
Hence option $\left(a\right)$ is correct.