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Question
Two stereo speakers are separated by a distance of 2.40 m. A person stands at a distance of 3.20 m directly in front of one of the speakers as shown in figure. Find the frequencies in the audible range (20-2000 Hz) for which the listener will hear a minimum sound intensity. Speed of sound in air = 320 m s−1.
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Solution
Given:
Distance between the two speakers d = 2.40 m
Speed of sound in air v = 320 ms−1
Frequency of the two stereo speakers f = ?
As shown in the figure, the path difference between the sound waves reaching the listener is given by :
\[∆ x = S_2 L - S_1 L\]
\[∆ x = \sqrt{(3 . 2 )^2 + (2 . 4 )^2} - 3 . 2\]
Wavelength of either sound wave:
\[= \left( \frac{320}{f} \right)\]
We know that destructive interference will occur if the path difference is an odd integral multiple of the wavelength.
\[\therefore ∆ x = \frac{(2n + 1)\lambda}{2}\]
So,
\[\sqrt{(3 . 2 )^2 + (2 . 4 )^2} - 3 . 2 = \frac{(2n + 1)}{2}\left( \frac{320}{f} \right)\]
\[ \Rightarrow \sqrt{16} - 3 . 2 = \frac{\left( 2n + 1 \right)}{2}\left( \frac{320}{f} \right)\]
\[ \Rightarrow 0 . 8 \times 2f = \left( 2n + 1 \right) \times 320\]
\[ \Rightarrow 1 . 6f = \left( 2n + 1 \right) \times 320\]
\[ \Rightarrow f = 200(2n + 1)\]
On putting the value of n = 1,2,3,...49, the person can hear in the audible region from 20 Hz to 2000 Hz.
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