Question

A simple harmonic progressive wave is represented by the equation y=8sin2π(0.1x−2t)y=8sin⁡2π(0.1x−2t) where x and y are in cm and t is in seconds. At any instant the phase difference between two particles separated by 2.0 cm in the x-direction is

Answer 1

Answer: The phase difference between two particles is $\phi =0.004\pi$.

Explanation:

A simple harmonic progressive wave is given in the problem as;

y=8sin2π(0.1x−2t)                                                                    

Rearrange the above equation.

y=8sin2π(-(2t-0.1x))

$y=8sin2\pi (2t-\frac{x}{10 })$                                        ....... (1)

Compare the above equation with the general equation.

$y=asin2\pi (nt-\frac{x}{\lambda })$                          ........ (2)

The value of lambda is 10 by comparing (1) and (2) equations.

It is given in the problem that at any instant the phase difference between two particles separated by 2.0 cm in the x-direction.  

The expression for the phase difference is;

$\phi =\frac{2\pi x}{\lambda }$

Put x=2 cm and ${\lambda }$ =10.

Calculate the phase difference by converting the distance 2 cm to m.

$\phi =\frac{2\pi (2 cm)}{10}$

$\phi =0.004\pi$

Therefore,  the phase difference between two particles is $\phi =0.004\pi$.