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 0.4s. The maximum speed of the vibrating particles in the medium is:

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

Answer 1

The correct option is A $\frac{5\pi }{2}cm/s$

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
$\nu =\frac{dy}{dt}=a\omega \sin (\omega t+kx)$
${\nu }_{max}=a\omega$

$a=1/2cm$
$=\frac{1}{2}\times \frac{2\pi }{0.4}=\frac{2\pi }{8}\times 10$
$=\frac{5\pi }{2}cm/s$