Question

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

• A. Amplitude of wave is less than $\frac{\lambda }{2\pi }$
• B. Amplitude of wave is greater than $\frac{\lambda }{2\pi }$
• C. Amplitude of wave is less than $\lambda$
• D. Amplitude of wave is greater than $\frac{\lambda }{\pi }$

Answer 1

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

Let, particle speed is $v_p$ and wave speed is $v$

So, according the given question $v_p<v$

Let, equation of wave is $y\left(x\right)=Asin(\omega t-kx)$

Now, $v=\frac{\omega }{k}$

The maximum particle speed is $(v_p{)}_m=A\omega$

So, paricle speed is always less than wave speed if $(v_p{)}_m<v$

$Or,A\omega <\frac{\omega }{k}$

$Or,A<\frac{1}{k}$

$Or,A<\frac{\lambda }{2\pi }$ $[Also,k=\frac{2\pi }{\lambda }]$

The correct answer is $OptionA:Amplitudeofwaveislessthan\frac{\lambda }{2\pi }$