Question

For the travelling harmonic wave y (x, t) = 2.0 cos $2\pi (10t-0.0080x+0.35)$ where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

(a) 4 m
(b) 0.5 m
(c) $\frac{\lambda }{2}$
(d) $\frac{3\lambda }{4}$

Answer 1

Equation for a travelling harmonic wave is given as:
y (x, t) = 2.0 cos $2\pi$ (10t - 0.0080x + 0.35)
$=2.0cos(20\pi t-0.016\pi x+0.70\pi )$
Where,
Propagation constant, $k=0.0160\pi$
Amplitude, a = 2 cm
Angular frequency, $\omega =20\pi rad/s$
Phase difference is given by the relation:
$\varphi =kx=\frac{2\pi }{\lambda }$
(a) For x =4m =400 cm
$\Phi =0.016\pi \times 400=6.4\pi rad$
(b) For 0.5 m = 50 cm
$\Phi =0.016\pi \times 50=0.8\pi rad$
(c) For $x=\frac{\lambda }{2}$
$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{2}=\pi rad$
(d) For $x=\frac{3\lambda }{4}$
$\varphi =\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}=1.5\pi rad$