Question

Two sources $S_1$ and $S_2$ separated by $2$ meters vibrate according to the equations $y_1=0.03\sin \pi t$ and $y_2=0.02\sin \pi t$ where $y_1$ and $y_2$ are in meters. They send out waves of velocity $1.5\frac{\text{m}}{\text{s}}$. The magnitude of the resultant motion of particle , collinear with $S_1$and $S_2$ and at the middle of $S_1{S}_2$ will be :

• A. $0.05\text{m}$
• B. $0.01\text{m}$
• C. $0.11\text{m}$
• D. $0.06\text{m}$

Answer 1

The correct option is A $0.05\text{m}$

$y_1=0.03\sin \pi t$ and $y_2=0.02\sin \pi t$

Each wave have the same period $=2$ seconds
Frequency, $n=\frac{1}{2}\text{Hz}$.

Since the wave velocity is $1.5\frac{m}{s}$, the wave length $\lambda =\frac{v}{n}=\frac{1.5}{\left(\frac{1}{2}\right)}=3\text{m}$

The path difference for a point $R$ at the center $S_1$ and $S_2=0$

Therefore, phase difference $=0$

The resultant amplitude , $A=(A_1^2+A_2^2+2A_1{A}_2\cos \varphi {)}^{\frac{1}{2}}=0.05\text{m}$