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प्रश्न
Two speakers S1 and S2, driven by the same amplifier, are placed at y = 1.0 m and y = −1.0 m(See figure). The speakers vibrate in phase at 600 Hz. A man stands at a point on the X-axis at a very large distance from the origin and starts moving parallel to the Y-axis. The speed of sound in air is 330 m s−1. (a) At what angle θ will the intensity of sound drop to a minimum for the first time? (b) At what angle will he hear a maximum of sound intensity for the first time? (c) If he continues to walk along the line, how many more can he hear?
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उत्तर
Given :
Frequency of source f = 600 Hz
Speed of sound in air v = 330 m/s
\[v = f\lambda\]
\[\therefore \lambda = \frac{v}{f}\]=\[\frac{330}{600} = 0 . 5 \text { mm }\]
Let the man travel a distance of \[\left( y \right)\]parallel to the y-axis and let \[\left( d \right)\] be the distance between the two speakers. The man is standing at a distance of \[\left( D \right)\]from the origin.
The path difference (x) between the two sound waves reaching the man is given by :
\[x = S_2 Q - S_1 Q = \frac{yd}{D}\]
Angle made by man with the origin :
\[\theta = \frac{y}{D}\]
Given:
d = 2 m
(a) For minimum intensity:
The destructive interference of sound (minimum intensity) takes place if the path difference is an odd integral
multiple of half of the wavelength.
\[\therefore x = (2n + 1)\left( \frac{\lambda}{2} \right)\]
\[\text { For } \left( n = 0 \right)\]
\[ \therefore \frac{yd}{D} = \frac{\lambda}{2}\]
\[\left( \because \theta = \frac{y}{D} \right)\]
\[ \therefore \theta d = \frac{\lambda}{2}\]
\[ \therefore \theta = \frac{\lambda}{2d} = \frac{0 . 55}{4} = 0 . 1375 \text{ rad }\]
\[ \Rightarrow \theta = 0 . 1375 \times (57 . 1)^\circ= 7 . 9^\circ\]
(b) For maximum intensity:
The constructive interference of sound (maximum intensity) takes place if the path difference is an integral multiple of the wavelength.
\[x = n\lambda\]
\[\text { For } \left( n = 1 \right): \]
\[ \Rightarrow \frac{yd}{D} = \lambda\]
\[ \Rightarrow \theta = \frac{\lambda}{d}\]
\[ \Rightarrow \theta = \frac{0 . 55}{2} = 0 . 275 \text { rad }\]
\[ \therefore \theta = 16^\circ\]
(c) The more number of maxima is given by the path difference :
\[\frac{yd}{D} = 2\lambda, 3\lambda, 4\lambda, . . . . . \]
\[ \Rightarrow \frac{y}{D} = \theta = 32^\circ, 64^\circ, 128^\circ\]
He will hear two more maxima at 32° and 64° because the maximum value of θ may be 90°.
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