Question

The equation of a simple harmonic progressive wave is given by
y = 8sin (0.628x – 12.56t)
where x and y are in cm and t is in second.
Find the phase difference in degrees between two particles at a distance of 2.0cm apart at any instant.

• A. 72
• B. 64
• C. 57
• D. 45

Answer 1

The correct option is A

72

$y=asin(\omega t-kx)$
The given equation is $y=8sin(0.628x-12.56t)$
​Comparing, we get
$a=8cm\to$ amplitude
$k=0.628rad{m}^{-1}\frac{2\pi }{\lambda }=0.628\lambda =\frac{2\pi }{0.628}=10cm\omega =12.56rad{s}^{-1}2\pi f=12.56f=\frac{12.56}{2\pi }=2H_{z}v=f\lambda =2\times 10=20cm{s}^{-1}$
Phase difference =$\Delta \varphi =\frac{2\pi }{\lambda }\times$ path difference
$\Delta \varphi =\frac{2\pi }{10}\times 2=\frac{2\pi }{5}={72}^{\circ }$