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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Waves and Sound

BITSAT 2015 · Physics · Waves and Sound

Two tuning forks with natural frequencies 340 Hz each move relative...

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 3Hz3 \mathrm{ Hz}. Find the speed of the tuning forks.

Options

  1. A1.5m/s1.5 \mathrm{ m} / \mathrm{s}
  2. B2m/s2 \mathrm{ m} / \mathrm{s}
  3. C1m/s1 \mathrm{ m} / \mathrm{s}
  4. D2.5m/s2.5 \mathrm{ m} / \mathrm{s}

Correct answer

A. 1.5m/s1.5 \mathrm{ m} / \mathrm{s}

Step-by-step solution

Let v=v= speed of sound and vS=v_{\mathrm{S}}= speed of tuning forks. Apparent frequency of fork moving towards the observer is n1=(vv−vs)n\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 n2=(vv+vs)n\mathrm{n}_{2}=\left(\frac{\mathrm{v}}{\mathrm{v}+\mathrm{v}_{\mathrm{s}}}\right) \mathrm{n}If fisthenumberofbeatsheardpersecond.thenf is the number of beats heard per second. then\mathrm{f}=\mathrm{n}_{1}-\mathrm{n}_{2} ⇒f=(vv−vs)n−(vv+vs)n⇒f=v(v+vs)−v(v−vs)\mathrmv2−vs2(n)⇒2vvsn\mathrmv2−vs2=f⇒2(vsv)=n=f{ if vs<Kv\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)}{\mathrmv^{2}-\mathrm{v}_{\mathrm{s}}^{2}}(\mathrm{n})\\ \Rightarrow \frac{2 \mathrm{vv}_{\mathrm{s}} \mathrm{n}}{\mathrmv^{2}-\mathrm{v}_{\mathrm{s}}^{2}}=\mathrm{f} \Rightarrow2\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}}=fv2n\frac{\mathrm{fv}}{2 \mathrm{n}} putting v=340 m/s,f=3,n=340 Hz we get, vs=340×33×340=1.5 m/s\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}$
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