Question

Two small loudspeakers A and B are driven by the same amplifier as shown in Fig and emit pure sinusoidal waves in phase. Speaker A is 1 m away as shown and speaker B is 2 m away from the amplifier. The microphone is 4 m away from the amplifier in transverse direction as indicated in the Figure. For what frequencies constructive interference will occur at P (microphone point) :

A. $1000{\text{Hz,2000Hz,}}.......$
B. $500{\text{Hz,1500Hz,}}.......$
C. $550{\text{Hz,1100Hz,}}.......$
D. $500{\text{Hz,}}1000{\text{Hz,1500Hz,}}.......$

Answer 1

Waves which are getting emitted from the two loudspeakers have the same frequency because they are in the same phase. So there will be interference at point P due to the waves coming from the two loudspeakers. Certain conditions to be followed in order to obtain constructive interference.

Formula used:
$\Delta x = n\lambda$
${H^2} = {M^2} + {N^2}$

Complete step by step answer:
The distance between point P and the amplifier is given as 4 meter. The distance between one loudspeaker A and point P is assumed to be ‘S’. The distance between other loudspeaker B and point P is assumed to be ‘L’.
First we should find out the values of ‘S’ and ‘L’ and then subtract one from the other to find out the path difference between the two waves.
From the pythagoras theorem we have ${H^2} = {M^2} + {N^2}$
Hence for the first triangle formed by first loudspeaker we have
${H^2} = {M^2} + {N^2}$
\eqalign{ & \Rightarrow {S^2} = {M^2} + {N^2} \cr & \Rightarrow {S^2} = {4^2} + {2^2} \cr & \Rightarrow {S^2} = 20 \cr & \Rightarrow S = \sqrt {20} \cr}
Now for the second triangle formed by the second loudspeaker we have
${H^2} = {M^2} + {N^2}$
\eqalign{ & \Rightarrow {L^2} = {M^2} + {N^2} \cr & \Rightarrow {L^2} = {4^2} + {1^2} \cr & \Rightarrow {L^2} = 17 \cr & \Rightarrow L = \sqrt {17} \cr}
The difference between them will be $S - L = 0.35$
In order to have constructive interference that path difference must be integral multiples of wavelength.
$\Delta x = n\lambda$
\eqalign{ & \Rightarrow S - L = n\lambda \cr & \Rightarrow 0.35 = n\lambda \cr}
But $\lambda = \dfrac{c}{f}$ where ‘c’ is the velocity of sound and assumed to be 350m/s and f is the frequency
$\lambda = \dfrac{c}{f}$
\eqalign{ & \Rightarrow \dfrac{{0.35}}{n} = \dfrac{c}{f} \cr & \Rightarrow \dfrac{{0.35}}{n} = \dfrac{{350}}{f} \cr & \Rightarrow f = 1000n \cr}
Where ‘n’ is any positive integer i.e n=1,2,3,………
So frequency must be multiples of thousand which is only satisfied by option A.
Hence option A will be the answer.

Note:
For the case of constructive interference i.e we will get the maximum output at point P we have this condition. There will be another condition for the case of destructive interference i.e for the minimum output there will be some other condition. Here phase difference is the function of space and phase difference is not constant.