Question

Two speakers S1 and S2, driven by the same amplifier, are placed at y = 1.0 m and y = −1.0 m. 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

Two speakers S1 and S2 , driven by the same amplifier, are placed at y = 1.0 m and y = -1.0 m (figure 16-E4). 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. (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 maxima can he hear ? Read more on Sarthaks.com - https://www.sarthaks.com/43465/two-speakers-s1-and-s2-driven-by-the-same-amplifier-are-placed-at-y-1-0-m-and-y-1-0-m-figure-16-e4

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Answer 2

Explanation:

Given data,  

$\mathrm{F}=600 \mathrm{hz}$

And v = 330 m/s

$\lambda=\frac{V}{f}$

$\lambda=\frac{330}{600}$

$\lambda=0.55 \mathrm{mm}$

Let OP = D

PQ = Y

$\theta=\frac{y}{R}$

Now the difference in path given by the

$x=\mathrm{S}_{2} \mathrm{Q}-\mathrm{S}_{1} \mathrm{Q}=\frac{y d}{D}$

Where  

D= 2 m  

The proof of $\frac{y d}{D}$ is Discussed in light wave interface

(a) For minimum intensity $x=\frac{(2 n+1) \lambda}{2}$

$\left.\frac{y d}{D}=\frac{\lambda}{2} \text { (for minimum } y, x=\frac{\lambda}{2}\right)$

$\frac{y}{D}=\theta=\frac{\lambda}{2}=\frac{0.55}{4}=0.1357 \mathrm{rad}=0.1357 \times 57.1^{\circ}=7.9^{\circ}$

(b) For minimum intensity $x=\frac{(2 n) \lambda}{2}$

$\frac{y d}{D}=\lambda$

$\frac{\lambda}{D}=\theta$

$\theta=\frac{0.55}{2}=0.275 \mathrm{rad}$

$\theta=16^{\circ}$

(c) For more maxima

$\frac{y d}{D}=2 \lambda, 3 \lambda, 4 \lambda \ldots \ldots$

$\frac{y}{D}=\theta=32^{\circ}, 64^{\circ}, 128^{\circ}$

Since he can hear a maximum value of 90 °, he can hear two more maximum values at 32 ° and 64 °.