Question

In a transverse wave the distance between a crest and neighbouring trough at the same instant is $4.0cm$ and the distance between a crest and trough at the same place is $1.0cm$. The next crest appears at the same place after a time interval of $0.4s$. The maximum speed of the vibrating particles in the medium is :

• A. $\frac{3\pi }{2}$ cm/s
• B. $\frac{5\pi }{2}$ cm/s
• C. $\frac{\pi }{2}$ cm/s
• D. $2\pi$ cm/s

Answer 1

Answer: (B)

Transverse wave equation
$y=a\sin (\omega t+kx)$
Next crest appears at time interval of $0.4s$
So, the time period $T=0.4$
$\therefore$ Angular Frequency $\omega =\frac{2\pi }{T}=\frac{2\pi }{0.4}$
$\therefore$ Speed of the particle
$v=\frac{dy}{dt}=a\omega \sin (\omega t+kx)$
$v_{\max }=a\omega$
$a=1/2cm$
$=\frac{1}{2}\times \frac{2\pi }{0.4}=\frac{2\pi }{8}\times 10$
$=\frac{5\pi }{2}cm/s$