Question

For the travelling harmonic wave $y(x,t)=2.0cos$ $2\pi$ (10t-0.0080 x+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 $x$

• A. $x=4m,\Delta \varphi =6.4\pi rad$
• B. $0.5m,\Delta \varphi =0.6\pi rad$
• C. $\lambda /2,\Delta \varphi =.6\pi rad$
• D. $3\lambda /4,\Delta \varphi =2.5\pi rad.$

Answer 1

The correct option is A $x=4m,\Delta \varphi =6.4\pi rad$

Equation for a travelling harmonic wave is given as:

$y(x,t)=2.0cos2\pi (10t-0.0080x+0.35)$

$=2.0cos(20\pi t-0.016\pi x+0.70\pi )$
Where,
Propagation constant, $k=0.0160\pi$
Amplitude, $a=2cm$
Angular frequency, $\omega =20\pi rad/s$

Phase difference is given by the relation:

$\varphi =kx=2\pi /\lambda$
(a) For $\Delta x=4m=400cm$

$\Delta \varphi =0.016\pi \times 400=6.4\pi rad$
(b) For $\Delta x=0.5m=50cm$

$\Delta \varphi =0.016\pi \times 50=0.8\pi rad$
(c) For $\Delta x=\lambda /2$

$\Delta \varphi =2\pi /\lambda \times \lambda /2=\pi rad$
(d) For $\Delta x=3\lambda /4$

$\Delta \varphi =2\pi /\lambda \times 3\lambda /4=1.5\pi rad$.