1. Minimum signal occurs at specific frequencies where sound waves from both speakers cancel out. For the lowest frequency, the path difference must be an odd multiple of half the wavelength (20 Hz in this case).
2. Subsequent minimum frequencies are found by multiplying the lowest frequency by factors of 3, 5, and so on (60 Hz, 100 Hz in this case).
3. Maximum signal occurs when the path difference is an even multiple of half the wavelength (30 Hz in this case).
4. Subsequent maximum frequencies are found by multiplying the lowest frequency by factors of 3, 5, and so on (90 Hz, 150 Hz in this case).
5. Remember these are theoretical frequencies and rounding is involved. Real-world factors might slightly modify the exact values.
## Solutions to two loudspeakers interference problem:
**a) Lowest frequency for minimum signal (destructive interference)**
To achieve destructive interference, the path difference between the sound waves from the two speakers must be an odd multiple of half the wavelength.
1. Calculate the path difference: 19.5m - 18.3m = 1.2m
2. Find the wavelength at the lowest frequency (20 Hz): Wavelength = Speed of sound / frequency = 343 m/s / 20 Hz = 17.15m
3. Solve for the lowest frequency with odd multiple half-wavelength path difference: 1.2m = (2n + 1)/2 * 17.15m
4. Rearrange and solve for n: n = (1.2m * 2 - 17.15m) / (17.15m) = -0.78
5. Since n must be an integer for a frequency, round up to n = 0. Therefore, the lowest frequency for minimum signal is fmin,1 = 20 Hz * (2 * 0 + 1)/2 = 20 Hz.
**b) Factor for next minimum frequency**
The next minimum frequency will occur when the path difference is another odd multiple of half the wavelength. So, you simply multiply the lowest frequency by the factor 2 + 1 = 3 to get the next frequency: fmin,2 = 20 Hz * 3 = 60 Hz.
**c) Factor for third minimum frequency**
Similarly, find the third minimum frequency by multiplying the lowest frequency by 2 + 3 = 5: fmin,3 = 20 Hz * 5 = 100 Hz.
**d) Lowest frequency for maximum signal (constructive interference)**
For constructive interference, the path difference must be an even multiple of half the wavelength. Repeat the calculations from part (a) but solve for an even multiple instead. You will get fmax,1 = 20 Hz * (2 * 1 + 1)/2 = 30 Hz.
**e) Factor for next maximum frequency**
For the next maximum frequency, multiply the lowest by 2 + 1 = 3: fmax,2 = 30 Hz * 3 = 90 Hz.
**f) Factor for third maximum frequency**
Finally, multiply the lowest by 2 + 3 = 5: fmax,3 = 30 Hz * 5 = 150 Hz.
**Therefore, the answers are:**
* (a) fmin,1 = 20 Hz
* (b) Factor = 3
* (c) Factor = 5
* (d) fmax,1 = 30 Hz
* (e) Factor = 3
* (f) Factor = 5
The probable question can be: Two loud speakers are located 3.35 m apart on an outdoor stage. A listener is 18.3 m from one and 19.5 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 Hz to 20 kHz).
(a) What is the lowest frequency fmin,1 that gives minimum signal (destructive interference) at the listener’s location?
(b) By what number must fmin,1 be multiplied to get the second lowest frequency fmin,2 that gives minimum signal
(c) the third lowest frequency fmin,3 that gives minimum signal?
(d) What is the lowest frequency fmax,1 that gives maximum signal (constructive interference) at the listener’s location?
(e) By what number must fmax,1 be multiplied to get the second lowest frequency fmax,2 that gives maximum signal ?
(f) the third lowest frequency fmax,3 that gives maximum signal?
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