Question

Two waves, each having a frequency of 100 Hz and a wavelength of 2.0 cm, are travelling in the same direction on a string. What is the phase difference between the waves (a) if the second wave was produced 0.015 s later than the first one at the same place, (b) if the two waves were produced at the same instant but the first one was produced a distance 4.0 cm behind the second one ? (c) If each of the waves has an amplitude of 2.0 mm, what would be the amplitudes of the resultant waves in part (a) and (b) ?

Answer 1

f = 100 Hz

$y=2cm=2\times {10}^{-2}m$

Wave speed, $v=fy=100\times 2\times {10}^{-2}m/s$

$=2m/s$

(a) In 0.015 sec 1st wave has travelled

$x=0.015\times 2$

$=0.03m$

= path difference

$\therefore$ Correpsponding phase diff.

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

$=\left\{\frac{2\pi }{(2\times {10}^{-2})}\right\}\times 0.03=3\pi$

(b) Path difference,

$x=4cm=0.04m$

$\Rightarrow \varphi =\left(\frac{2\pi }{\lambda }\right)x$

$=\left\{\right(\frac{2\pi }{(2\times {10}^{-2})})\times 0.04\}$

$=4\pi$
(c) The waves have same frequency, same wavelength and same amplitude.

Let $y_1=rsinwt$

$y_2=rsin(wt+\varphi )$

$\Rightarrow y=y_1+y_2$

$=r[sinwt+(wt+\varphi \left)\right]$

$=2rsin(wt+\frac{\varphi }{2})cos\left(\frac{\varphi }{2}\right)$

$\therefore$ Resultant amplitude $=2rcos\frac{\varphi }{2}$

So, when $\varphi =3x$

$r=2\times {10}^{-3}m$

$R_{res}=2\times (2\times {10}^{-3})cos\left(\frac{3\pi }{2}\right)$

Again, when $\varphi =4\pi$

$R_{res}=2\times (2\times {10}^{-3})cos\left(\frac{4\pi }{2}\right)$

$=4\times {10}^{-3}\times \left(1\right)m$

$=4mm$