Question

For the travelling harmonic wave
$y(x, t) = 2\, cos2\pi(10t - 0.008x + 0.35)$ where $x$ and $y$ are in cm and $s$ is in $s$. The phase difference between oscillatory motion 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)

The given equation is
$y = 2 \,cos2\pi (10t - 0.008x + 0.35)$
$y = 2\, cos(20\pi t - 0.016\,\pi x + 0 .77\pi) \quad ... (i)$
The standard equation of travelling harmonic wave is
$y = acos\left(\omega t-kx+\phi\right)\quad... \left(ii\right)$
Comparing $\left(i\right)$ and $\left(ii\right)$, 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
or $\Delta\phi = \frac{2\pi }{\lambda } \Delta x$
When, $\Delta x = 0.5 \,m = 50 \,cm$
$\therefore \Delta\phi = 2\pi \times 0.008 \times 50$
$= 0.8\pi\,rad$