Question

Two speakers separated by distance $d_{1} = 2.00 m$ are in phase. Assume the amplitudes of the sound waves from the speakers are approximately the same at the listener's ear at distance $d_{2} = 4.00 m$ directly in front of one speaker. Consider the full audible range for normal hearing. $20 Hz$ to $20 kHz$ .

(a) What is the lowest frequency $f_{\min . 1}$ that gives the minimum signal (destructive interference) at the listener's ear?

By what number must $f_{\min . 1}$ be multiplied to get

(b) the second lowest frequency $f_{\min . 2}$ that gives the minimum signal

(c) the third lowest frequency $f_{\min . 3}$ that gives the minimum signal?

(d) what is the lowest frequency $f_{\max . 1}$ that gives the maximum signal (constructive interference) at the listener's ear?

By what number must $f_{\max . 1}$ be multiplied to get

(e) the second lowest frequency $f_{\max . 2}$ that gives the maximum signal and

(f) the third lowest frequency $f_{\max . 3}$ that gives the maximum signal?

Accepted Answer

Given,

The distance between the two speakers is, $d_{1} = 2.00 m$

The distance between the second speaker and listener is, $d_{2} = 4.00 m$

The path traveled by wave from first speaker to the listener is, $S_{1} E = x_{1} = \sqrt{{d_{1}}^{2} + {d_{2}}^{2}} = \sqrt{2^{2} + 4^{2}} = 2 \sqrt{5} m$

The path traveled by wave from second speaker to the listener is, $S_{2} E = x_{2} = d_{2} = 4.00 m$

Now, the path difference between the two sound waves received by the listener is, $S_{1} E - S_{2} E = 2 \sqrt{5} - 4 = 0.47 m$

The full audible range for normal hearing is, $20 Hz$ to $20 kHz$ .

(a) The condition of destructive interference between two sound waves is given by,

Path difference, $x_{1} - x_{2} = (2 n - 1) \frac{λ}{2}$ , where, $n = 1, 2, 3 . . . . . .$

$\Rightarrow λ = \frac{2 (x_{1} - x_{2})}{(2 n - 1)} . . . (1)$

Now, the frequency of sound wave is given by,

$f_{m i n} = \frac{v}{λ} = \frac{v (2 n - 1)}{2 (x_{1} - x_{2})} . . . (2)$

$f_{m i n} = \frac{v (2 n - 1)}{2 \times 0.47} = \frac{v (2 n - 1)}{0.94}$

The lowest frequency $f_{\min . 1}$ that gives minimum signal (destructive interference) at the listener's ear, when $n = 1,$ $\Rightarrow f_{\min . 1} = \frac{v}{λ} = \frac{343}{0.94} = 364.89 Hz$

(b) For $n = 2,$ Equation $(2) \Rightarrow f_{\min . 2} = 3 \times f_{\min . 1}$

Therefore, to get second lowest frequency $f_{m i n . 2}$ , we must multiply $f_{m i n . 1}$ by $3$ .

(c) For $n = 3,$ Equation $(2) \Rightarrow f_{\min . 3} = 5 \times f_{\min . 1}$ ,

Therefore, to get second lowest frequency $f_{\min . 3}$ , we must multiply $f_{m i n . 1}$ by $5$ .

d) The condition for constructive interference between two sound waves is given by,

Path difference, $x_{1} - x_{2} = n λ$ , where, $n = 1, 2, 3 . . . . .$

$\Rightarrow λ = \frac{x_{1} - x_{2}}{n} . . . (3)$

Frequency for constructive interference,

$f_{m a x} = \frac{n v}{x_{1} - x_{2}} . . . (4)$

The lowest frequency $f_{\max . 1}$ that gives maximum signal at the listener's ear,

$n = 1, f_{\max . 1} = \frac{343}{0.47} = 729.79 Hz$

(e) For $n = 2,$ Equation $(4) \Rightarrow f_{m a x . 2} = 2 \times f_{m a x . 1}$

Therefore, to get second lowest frequency $f_{m a x . 2}$ , we must multiply $f_{m a x . 1}$ by $2$ .

(f) For $n = 3,$ Equation $(2) \Rightarrow f_{m a x . 3} = 3 \times f_{m a x . 1}$

Therefore, to get second lowest frequency $f_{m a x . 3}$ , we must multiply $f_{m a x . 1}$ by $3$ .

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