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Sound Waves — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsSound Waves5 Marks
Question
A small source of sound S of frequency 500Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6m. The string just remains tight wher the source is at the highest point.
  1. An observer is located in the same vertical plane at a large distance and at the same height as the centre of the circle figure. The speed of sound in air = 330m/s and g = 10m/s 2. Find the maximum frequency heard by the observer.
  2. An observer is situated at a large distance vertically above the centre of the circle. Find the frequencies heard by the observer corresponding to the sound emitted by the source when it is at the same height as the centre.
✓

Answer

Given that, r = 1.6m, f = 500Hz, u = 330m/s
  1. At A, velocity of the particle is given by,
$\text{v}_\text{A}=\sqrt{\text{rg}}=\sqrt{1.6\times10}=4\text{m/s}$
and at C, $\text{v}_\text{A}=\sqrt{\text{rg}}=\sqrt{1.6\times10}=4\text{m/s}$
So, maximum frequency at C,
$\text{f}'_\text{c}=\frac{\text{u}}{\text{u}-\text{v}_\text{s}}\text{f}=\frac{330}{330-8.9}\times500=513.85\text{Hz}.$
Similarly, maximum frequency at A is given by $\text{f}'_\text{A}=\frac{\text{u}}{\text{u}-(-\text{v}_\text{s})}\text{f}=\frac{330}{330+4}(500)=494\text{Hz}.$
  1. Velocity at $\text{B}=\sqrt{3\text{rg}}=\sqrt{3\times1.6\times10}=6.92\text{m/s}$

So, frequency at B is given by, $\text{f}'_\text{B}=\frac{\text{u}}{\text{u}+\text{v}_\text{s}}\times\text{f}=\frac{330}{330+6.92}\times500=490\text{Hz}$
and frequency at D is given by,
$\text{f}_\text{D}=\frac{\text{u}}{\text{u}-\text{v}_\text{s}}\times\text{f}$
$=\frac{330}{330-6.92}\times500$
$=511\text{Hz}$

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