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Question
A source of sound emitting a 1200 Hz note travels along a straight line at a speed of 170 m s−1. A detector is placed at a distance 200 m from the line of motion of the source. (a) Find the frequency of sound receive by the detector at the instant when the source gets closest to it. (b) Find the distance between the source and the detector at the instant in detects the frequency 1200 Hz. Velocity of sound in air = 340 m s−1.
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Solution
Given:
Velocity of the source \[v_s\] = 170 m/s
Frequency of the source \[f_0\] = 1200Hz
(a)
As shown in the figure,
the time taken by the sound to reach the listener is the same as the time taken by the sound to reach the point of intersection.
\[\frac{y}{170} = \frac{\sqrt{{200}^2 + y^2}}{340}\]
\[ \Rightarrow \left( 2y \right)^2 = \left( 200 \right)^2 + \left( y \right)^2 \]
\[ \Rightarrow 4 y^2 - \left( y \right)^2 = \left( 200 \right)^2 \]
\[ \Rightarrow 3 \left( y \right)^2 = \left( 200 \right)^2 \]
\[ \Rightarrow y = \frac{200}{\sqrt{3}}\]
Frequency of source will be :
\[v_s \cos\theta\] = \[170 . \frac{y}{\sqrt{{200}^2 + y^2}} = 170 \times \frac{1}{2} = 85\]
The frequency of sound \[\left( f \right)\] heard by the detector is given by :
\[f = \frac{v}{v - v_s \cos v} \times f_0 \]
\[ \Rightarrow f = \frac{340}{340 - 170 \times \frac{1}{2}} \times 1200\]
\[ \Rightarrow f = 1600 \text{ Hz }\]
(b) The detector will detect a frequency of 1200 Hz at a minimum distance.
\[\frac{200}{340} = \frac{x}{170}\]
\[ \Rightarrow x = 100 \text { m }\]
∴ Distance
\[= \sqrt{\left( 200 \right)^2 + x^2} = \sqrt{\left( 200 \right)^2 + \left( 100 \right)^2}\]
\[ = 224 \text{ m }\]
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