A wave of frequency 500 Hz has wave velocity of 350 m/s. 1)Find the distance between the two points which has 60^o out of phase. 2)Find the phase difference between two displacements at certain point of time 10^-3s apart?

Jun 1, 2017

Given velocity of the wave $v = 350 m \text{/} s$

frequency of the wave $n = 500 H z$

So wave length of the wave $\lambda = \frac{v}{n} = \frac{350}{500} m = 0.7 m$

1) We are to find the distance between the two points which has ${60}^{\circ}$ out of phase i.e the phase difference is $\phi = {60}^{\circ} = \frac{\pi}{3} r a d$

As we know that for path difference $\lambda$ there is phase difference $2 \pi$,

we can say , if $\phi$ is the phase difference for path difference $x$,then

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

$x = \frac{\phi \lambda}{2 \pi} = \frac{\pi}{3} \times \frac{0.7}{2 \pi} = \frac{0.7}{6} m \approx 0.116 m$

2) Now in $t = {10}^{-} 3 s$ the wave moves

$v \times t = 350 \times {10}^{-} 3 = 0.35 m$

So here path difference $x = 0.35 m$

So by the relation $\phi = \frac{2 \pi x}{\lambda}$, the phase difference for $x = 0.35 m$ becomes
$\phi = \frac{2 \pi \times 0.35}{0.7} = \pi$,