Question

Equation of progressive wave is given by, $y=4sin[\pi (\frac{t}{5}-\frac{x}{9})+\frac{\pi }{6}],$ where x and y are in metre. Then

• A. $v=5m/s$
• B. $\lambda =18m$
• C. $A=0.04m$
• D. $f=50Hz$

Answer 1

Answer: (B)

$\lambda =18m$

For a wave $y=Asin(\omega t-kx+\varphi )$

$k=\frac{2\pi }{\lambda }$

$\omega =2\pi \nu$

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

Here, $\omega =\frac{\pi }{5}$

$k=\frac{\pi }{9}$

Thus $\lambda =\frac{2\pi }{\pi /9}=18m$

$\nu =\frac{\omega }{2\pi }$

$=\frac{\pi /5}{2\pi }$

$=0.1Hz$

$v=\frac{\pi /5}{\pi /9}$

$=9/5m/s$

$A=4m$

Thus correct answer is option B.