Question

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?

Answer 1

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°.