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Two loud speakers are located \(3.35 \mathrm{~m}\) apart on an outdoor stage. A listener is \(18.3 \mathrm{~m}\) from one and \(19.5 \mathrm{~m}\) from the other. During the sound check, a signal generator drives the two speakers in phase with the same amplitude and frequency. The transmitted frequency is swept through the audible range \((20 \mathrm{~Hz}\) to \(20 \mathrm{kHz}) .\) (a) What is the lowest frequency \(f_{\min , 1}\) that gives minimum signal (destructive interference) at the listener's location? By what number must \(f_{\min , 1}\) be multiplied to get (b) the second lowest frequency \(f_{\min 2}\) that gives minimum signal and (c) the third lowest frequency \(f_{\min , 3}\) that gives minimum signal? (d) What is the lowest frequency \(f_{\max , 1}\) that gives maximum signal (constructive interference) at the listener's location? By what number must \(f_{\max , 1}\) be multiplied to get (e) the second lowest frequency \(f_{\max , 2}\) that gives maximum signal and (f) the third lowest frequency \(f_{\max , 3}\) that gives maximum signal?

Step by step solution

01

Understanding the Concept of Interference

For destructive interference, the path difference must be an odd multiple of half the wavelength, i.e., \((2n+1)\frac{\lambda}{2}\) where \(n = 0, 1, 2, \ldots\). For constructive interference, the path difference is an integer multiple of the wavelength \(m\lambda\) where \(m = 0, 1, 2, \ldots\).

02

Calculate Path Difference

Find the path difference \(\Delta L\) using the distances from the listener to each speaker: \(\Delta L = 19.5 \text{m} - 18.3 \text{m} = 1.2 \text{m}\).

03

Frequency for Minimum Signal (Destructive Interference)

For the lowest frequency giving destructive interference: \(\Delta L = (2n+1)\frac{\lambda}{2}, n=0\). Using the speed of sound \(c = 343\text{ m/s}\), \(\lambda = \frac{c}{f}\). Set \(\Delta L = \frac{\lambda}{2}\): \[1.2 = \frac{343}{2f_{\min, 1}}\].Solve for \(f_{\min, 1}\): \[f_{\min, 1} = \frac{343}{2 \times 1.2} \approx 142.92 \text{Hz}\].

04

Further Destructive Interference Frequencies

For consecutive destructive frequencies: \(f_{\min, 2} = 3f_{\min, 1}\) and \(f_{\min, 3} = 5f_{\min, 1}\). Thus, the multipliers are 3 and 5.

05

Frequency for Maximum Signal (Constructive Interference)

Set \(\Delta L = m\lambda, m=1\): \[1.2 = \frac{343}{f_{\max, 1}}\].Solve for \(f_{\max, 1}\): \[f_{\max, 1} = \frac{343}{1.2} \approx 285.83 \text{Hz}\].

06

Further Constructive Interference Frequencies

For consecutive constructive frequencies: \(f_{\max, 2} = 2f_{\max, 1}\) and \(f_{\max, 3} = 3f_{\max, 1}\). Thus, the multipliers are 2 and 3.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Destructive Interference

Destructive interference occurs when two waves meet and effectively cancel each other out. This happens when the path difference between the two waves is such that they are out of phase. For destructive interference to occur, the path difference must be an odd multiple of half the wavelength, given by the formula \[(2n+1)\frac{\lambda}{2}\]where \(n\) is an integer (0, 1, 2, ...).

In the given problem, the path difference between the two waves is calculated as 1.2 meters. To determine the lowest frequency that results in destructive interference, the path difference is equated to half the wavelength. By using the speed of sound (343 m/s), you can solve for the frequency:
\[f_{\min, 1} = \frac{343}{2 \times 1.2} \approx 142.92 \text{ Hz}\]This frequency creates the first instance of destructive interference at the listener's location.

Constructive Interference

Constructive interference happens when two waves meet in phase and reinforce each other. The resulting wave amplitude is larger, leading to a maximum signal. For constructive interference, the path difference corresponds to a whole number multiple of the wavelength:\[m\lambda\]where \(m\) is an integer (0, 1, 2, ...).

In this problem, the path difference between the sound waves from the two speakers is also 1.2 meters. To find the lowest frequency for constructive interference, set the path difference equal to a full wavelength and solve for the frequency:
\[f_{\max, 1} = \frac{343}{1.2} \approx 285.83 \text{ Hz}\]This frequency gives the listener a maximum signal as the waves construct upon each other in phase.

Audible Frequency Range

The human ear is capable of hearing a wide range of frequencies, from 20 Hz to 20 kHz. This range is known as the audible frequency range. In this exercise, the sound frequency transmitted by the speakers is swept across this range. It's essential in practical applications, such as testing audio equipment or acoustics in architecture.

In the context of wave interference, only certain frequencies within this range will lead to noticeable destructive or constructive interference, given the listener's fixed position. Identifying these frequencies requires calculating based on the speaker's distance from the listener and the speed of sound.

Path Difference

The path difference is crucial in determining how waves interfere at a listener's location. It is the difference in distance traveled by two waves before they meet at a point. Path difference determines whether the interference will be constructive or destructive, based on relative phase differences of the waves.

In this scenario, the listener is 18.3 m from one speaker and 19.5 m from the other, resulting in a path difference of:\[\Delta L = 19.5 \text{ m} - 18.3 \text{ m} = 1.2 \text{ m}\]This constant path difference is used to find frequencies for which the interfering sound waves either cancel out (destructive) or strengthen (constructive) each other, determining the quality of the sound heard by the listener.