Search any question & find its solution
Question:
Answered & Verified by Expert
Two tuning forks with natural frequencies 340 Hz each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards the observer at the same speed. The observer hears beats offrequency $3 \mathrm{~Hz}$. Find the speed of the tuning forks.
Options:
Solution:
1361 Upvotes
Verified Answer
The correct answer is:
$1.5 \mathrm{~m} / \mathrm{s}$
Let $v=$ speed of sound and $v_{\mathrm{S}}=$ speed of tuning forks. Apparent frequency of fork moving towards the observer is
$$
\mathrm{n}_{1}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}
$$
Apparent frequency of the fork moving away from the observer is
$$
\mathrm{n}_{2}=\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}
$$
If $f$ is the number of beats heard per second. then $\mathrm{f}=\mathrm{n}_{1}-\mathrm{n}_{2}$
$$
\begin{array}{l}
\Rightarrow \mathrm{f}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}-\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n} \\
\Rightarrow \mathrm{f}=\frac{\mathrm{v}\left(\mathrm{v}+\mathrm{v}_{\mathrm{s}}\right)-\mathrm{v}\left(\mathrm{v}-\mathrm{v}_{\mathrm{s}}\right)}{\mathrm{v}^{2}-\mathrm{v}_{\mathrm{s}}^{2}}(\mathrm{n}) \\
\Rightarrow \frac{2 \mathrm{vv}_{\mathrm{s}} \mathrm{n}}{\mathrm{v}^{2}-\mathrm{v}_{\mathrm{s}}^{2}}=\mathrm{f} \Rightarrow 2\left(\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}}\right)=\mathrm{n}=\mathrm{f}\left\{\right.\text { if } \mathrm{v}_{\mathrm{s}}<\mathrm{K}v
\end{array}
$$
$$
\Rightarrow v_{\mathrm{s}}=\frac{\mathrm{fv}}{2 \mathrm{n}}
$$
$$
\begin{array}{l}
\text { putting } \mathrm{v}=340 \mathrm{~m} / \mathrm{s}, \mathrm{f}=3, \mathrm{n}=340 \mathrm{~Hz} \text { we get, } \\
\mathrm{v}_{\mathrm{s}}=\frac{340 \times 3}{3 \times 340}=1.5 \mathrm{~m} / \mathrm{s}
\end{array}
$$
$$
\mathrm{n}_{1}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}
$$
Apparent frequency of the fork moving away from the observer is
$$
\mathrm{n}_{2}=\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}
$$
If $f$ is the number of beats heard per second. then $\mathrm{f}=\mathrm{n}_{1}-\mathrm{n}_{2}$
$$
\begin{array}{l}
\Rightarrow \mathrm{f}=\left(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}-\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n} \\
\Rightarrow \mathrm{f}=\frac{\mathrm{v}\left(\mathrm{v}+\mathrm{v}_{\mathrm{s}}\right)-\mathrm{v}\left(\mathrm{v}-\mathrm{v}_{\mathrm{s}}\right)}{\mathrm{v}^{2}-\mathrm{v}_{\mathrm{s}}^{2}}(\mathrm{n}) \\
\Rightarrow \frac{2 \mathrm{vv}_{\mathrm{s}} \mathrm{n}}{\mathrm{v}^{2}-\mathrm{v}_{\mathrm{s}}^{2}}=\mathrm{f} \Rightarrow 2\left(\frac{\mathrm{v}_{\mathrm{s}}}{\mathrm{v}}\right)=\mathrm{n}=\mathrm{f}\left\{\right.\text { if } \mathrm{v}_{\mathrm{s}}<\mathrm{K}v
\end{array}
$$
$$
\Rightarrow v_{\mathrm{s}}=\frac{\mathrm{fv}}{2 \mathrm{n}}
$$
$$
\begin{array}{l}
\text { putting } \mathrm{v}=340 \mathrm{~m} / \mathrm{s}, \mathrm{f}=3, \mathrm{n}=340 \mathrm{~Hz} \text { we get, } \\
\mathrm{v}_{\mathrm{s}}=\frac{340 \times 3}{3 \times 340}=1.5 \mathrm{~m} / \mathrm{s}
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.