Question

Two sound sources shown in the figure vibrate in phase. By moving $S_1$ along $P{S}_1$ consecutive minima are heard when $L_1-L_2$ has values, $20\text{cm},60\text{cm}$ and $100\text{cm}$. If the frequency of sound source is $\frac{1700}{n}\text{Hz}$. Then find the value of $n$ : [Speed of sound is $340\text{m/s}$ ]

Answer 1

Path difference between the sound waves along $S_{1}P$ and $S_{2}P$ is given by $|S_{1}P-S_{2}P|=|L_1-L_2|$

To hear a minima, the path difference must be equal to $(n+1)\frac{\lambda }{2},n=0,1,2,3,\mathrm{4....}$.

Thus, $|S_{1}P-S_{2}P|=(n+1)\frac{\lambda }{2}$

Therefore, $|L_1-L_2|=(n+1)\frac{\lambda }{2}$

For, $|L_1-L_2|=20\text{cm}$ , let us consider $n=0$

$\frac{\lambda }{2}=0.2$

$\Rightarrow \lambda =0.4\text{m}$

Similarly, for $|L_1-L_2|=40\text{cm}$ ,consider $n=1$ and for $|L_1-L_2|=60\text{cm}$ , consider $n=2$.

Frequency of sound source $f=\frac{v}{\lambda }=\frac{340}{0.4}=\frac{1700}{2}$

Comparing with the data given in the question we get, $n=2$

Accepted answer: $2$