Question

In a transverse wave the distance between a crest and neighbouring trough at the same instant is $4.0\, cm$ and the distance between a crest and trough at the same place is $1.0\, cm$. The next crest appears at the same place after a time interval of $0.4\,s$. 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
$y=a \sin (\omega t+k x)$
Next crest appears at time interval of $0.4 s$
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
$v=\frac{d y}{d t}=a \omega \sin (\omega t+k x)$
$v_{\max }=a \omega$
$a=1 / 2 \,cm$
$=\frac{1}{2} \times \frac{2 \pi}{0.4}=\frac{2 \pi}{8} \times 10$
$=\frac{5 \pi}{2} \,cm / s$