Question

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

The correct option is D $0.8\pi$ rad

The given equation is
$y=2cos2\pi (10t-0,008+0.35)$
$y=2cos(20\pi t-0.016\pi x+0.7\pi$ .....(i)
The standard equation of travelling harmonic wave is
$y=acos(\omega t-kx+\varphi )$ .....(ii)
Comparing (i) and (ii), we get
$k=0.016,\frac{2\pi }{\lambda }$ or $\lambda =\frac{1}{0.008}cm$
Phase different= $\frac{2\pi }{\lambda }\times$ Path difference or $△\varphi =\frac{2\pi }{\lambda }△x$
When,$△x=0.5m=50cm$
$\therefore △\varphi =2\pi \times 0.008\times 50=0.8\pi$ rad