Question

Sources separated by $20\text{m}$ vibrate according to the equation $y_1=0.06\sin \pi t$ and $y_2=0.02\sin \pi t.$ They send out waves along a rod with speed $3\text{m/s}$. What is the equation of motion of a particle $12\text{m}$ from the first source and $8\text{m}$ from the second? Given $y_1,y_2$ are in metres.

• A. $0.0173\sin \pi t-0.05\cos \pi t$
• B. $0.0173\sin \pi t-0.017\cos \pi t$
• C. $0.05\sin \pi t-0.0173\cos \pi t$
• D. $0.05\sin \pi t-0.5\cos \pi t$

Answer 1

The correct option is C $0.05\sin \pi t-0.0173\cos \pi t$

Given that
$y_1=0.06\sin \pi t\&y_2=0.02\sin \pi t$
So, $\omega =\pi$
Also, $v=3\text{m/s}$ [given]
Angular wave number $\left(k\right)=\frac{\omega }{v}=\frac{\pi }{3}\text{rad/m}$
Now, equation of disturbance due to source 1
$y_1=0.06\sin (\pi t-kx)=0.06\sin (\pi t-\pi /3\times 12)$
$\Rightarrow y=0.06\sin (\pi t-4\pi )$
Similarly,
$y_2=0.02(\pi t-k{x}^{\prime })=0.02(\pi t-\frac{\pi }{3}\times 8)$
$\Rightarrow y_2=0.02\sin (\pi t-\frac{8\pi }{3})$
Now, applying principle of superposition.
$y=y_1+y_2$
$=0.06\sin (\pi t-4\pi )+0.02\sin (\pi t-\frac{8\pi }{3})$
Using $\sin (A-B)=\sin A\cos B-\cos A\sin B$,
$=0.06\sin \left(\pi t\right)+0.02[\sin \pi t\cos \frac{8\pi }{3}-\cos \pi t\sin \frac{8\pi }{3}]$
$\sin \frac{8\pi }{3}=\sin (2\pi +\frac{2\pi }{3})=\sin \frac{2\pi }{3}$
and $\cos \frac{8\pi }{3}=\cos (2\pi +\frac{2\pi }{3})=\cos \frac{2\pi }{3}$
Substituting in equation,
$y=0.06\sin \pi t-0.01\sin \pi t-0.0173\cos \pi t]$
i.e $y=0.05\sin \pi t-0.0173\cos \pi t$