Question

For the travelling harmonic wave
$y(x,t)=2cos2\pi (10t-0.008x+0.35)$ 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 $0.5m$ 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
$y=2cos2\pi (10t-0.008x+0.35)$
$y=2cos(20\pi t-0.016\pi x+0.77\pi )...(i)$
The standard equation of travelling harmonic wave is
$y=acos\left(\omega t-kx+\varphi \right)...\left(ii\right)$
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 $\Delta \varphi =\frac{2\pi }{\lambda }\Delta x$
When, $\Delta x=0.5m=50cm$
$\therefore \Delta \varphi =2\pi \times 0.008\times 50$
$=0.8\pi rad$