Question

When a wave travels in a medium, the particle displacement is given by $y=a\sin 2\pi (bt-cx),$ where $a,b$ and $c$ are constants. The maximum particle velocity will be twice the wave velocity if

• A. $c=\frac{1}{\pi a}$
• B. $c=\pi a$
• C. $b=ac$
• D. $b=\frac{1}{ac}$

Answer 1

The correct option is A $c=\frac{1}{\pi a}$

General expression for a harmonic progressive wave is given by $y=A\sin (\omega t-kx)$.

Comparing with given equation $y=a\sin 2\pi (bt-cx)$.

Here, $2\pi f=\omega =2\pi b$ or $f=b$
and $k=\frac{2\pi }{\lambda }=2\pi c$ or $\frac{1}{\lambda }=c$

$\therefore$ Velocity of the wave $v_w=f\lambda =b\times \frac{1}{c}=\frac{b}{c}\dots \left(1\right)$

Particle velocity is given by
$v_P=\frac{dy}{dt}=a\times 2\pi b\cos 2\pi (bt-cx)=a\omega \cos (\omega t-cx)$
& Maximum particle velocity $v_{max}=a\omega =a\times 2\pi b=2\pi ab\dots \left(2\right)$

Given $v_{max}=2v_w$
$\Rightarrow 2\times \frac{b}{c}=2\pi ab$
$\Rightarrow 2\pi a=\frac{2}{c}$ or $c=\frac{1}{\pi a}$
Hence, the correct answer is option (a).

Why this question? Caution: Here $c$ is the symbol given for $\frac{1}{\lambda }$ and not the velocity of wave.