Question

For the travelling harmonic wave y(x, t) = 2 cos 2p(10t – 0.008x + 0.35) where x and y are in cm and t is in s. The phase difference between oscillatory motions of two points separated by a distance of 0.5 m is

• A. 0.2$\pi$rad
• B. 0.4$\pi$ rad
• C. 0.6$\pi$ rad
• D. 0.8$\pi$ rad

Answer 1

Answer: (D)

0.8$\pi$ rad

Answer 2

The given equations is

$y = 2\cos 2\pi(10t - 0.008x + 0.35)$

$y = 2\cos(20\pi t - 0.016\pi x + 0.7\pi)$ … (i)

The standard equations of travelling harmonic wave is

$y = a\cos(\omega t - kx + \varphi)$ ….. (ii)

Comparing (i) and (ii), we get

$k = 0.016\pi$

$\frac{2\pi}{\lambda} = 0.016\pi$

Or $\lambda = \frac{1}{0.008}cm$

Phase difference $= \frac{2\pi}{\lambda} \times$Path difference

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

Where, $\Delta x = 0.5m = 50cm$

$\therefore\Delta\varphi = 2\pi \times 0.008 \times 50 = 0.8\pi$