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Question
A small source of sound S of frequency 500 Hz is attached to the end of a light string and is whirled in a vertical circle of radius 1.6 m. The string just remains tight when the source is at the highest point. (a) An observer is located in the same vertical plane at a large distance and at the same height as the centre of the circle. The speed of sound in air = 330 m s−1 and g = 10 m s−2. Find the maximum frequency heard by the observer. (b) An observer is situated at a large distance vertically above the centre of the circle. Find the frequency heard by the observer corresponding to the sound emitted by the source when it is at the same height as the centre.
Solution
Given:
Frequency of sound emitted by the source \[f_0\] = 500 Hz
Velocity of sound in air v = 330 ms-1
Radius of the circle r = 1.6 m
Frequency of sound heard by the observer v, = ?
(a)
Velocity of source at highest point of the circle A is given by :
\[v_s = \sqrt{rg}\]=\[\sqrt{10 \times 1 . 6} = 4 \text { m/s }\]
Velocity of sound at C is \[v_c = \sqrt{5rg} = \sqrt{5 \times 1 . 6 \times 10} = 8 . 9 \text { m/s }\]
The frequency of sound heard by the observer when the source is at point C :
\[f_C = \frac{v}{v - v_s} \times f_0\]
Substituting the values, we get :
\[f_C = \frac{330}{330 - 8 . 9} \times 500\]
\[ \Rightarrow f_C = 513 . 85 \text { Hz } \approx 514 \text { Hz }\]
Frequency of sound observed by the observer when the source is at point A :
\[f_A = \frac{v}{v + v_s} \times n_0 \]
\[ = \frac{300}{300 + 4} \times 500 = 494 \text { Hz }\]
Therefore, maximum frequency heard by the observer is 514 Hz.
(b) Velocity at B is given by :
\[v_B = \sqrt{3rg} = \sqrt{3 \times 1 . 6 \times 10} = 6 . 92 \text { m/s }\]
Frequency at B \[\left( f_B \right)\] will be :
\[f_B = \frac{v}{v + v_s} \times f_0 \]
\[ = \frac{330}{330 + 6 . 92} \times 500 = 490 \text { Hz }\]
Frequency at D \[\left( f_D \right)\] will be :
\[f_D = \frac{v}{v - v_s} \times f_0 \]
\[ = \frac{330}{330 - 6 . 92} \times 500 = 511 \text{ Hz }\]
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