Calculate the wavelength in air at 20 C for sounds in the maximum range of human

Question

Calculate the wavelength in air at 20 C for sounds in the maximum range of human bearing 20 Hz to 20,000 Hz (problem 3) What is the intensity of a sound at the pain level of 120 dB? Compare it to that of a whisper at 20 dB (problem 14) Two stereo amplifiers are rated at 100 W and 150 W, How much louder will the more powerful amplifier be when both are producing sound at their maximum level (problem is) A If the sound intensity at a distance 50 cm from a person speaking in normal conversation is 3 times 10^-6 W/m^2. Assuming that the sound spreads roughly uniformly over s sphere centered on the mouth, (problem 23) Estimate the power output of that sound How many people would it take to produce a total sound output of 75 W of ordinary conversation? The predominant frequency of a certain fire truck's siren is 1350 Hz when at rest. What frequency do you detect if you move with a speed 30 m/s? (problem 61) Toward the fire truck, and Away from it A bat at rest sends out ultrasonic sound waves at 50 kHz and receives them returned from an object moving directly away from it at 30 m/s. What is the received sound frequency? (problem 62) A police car sounding a siren with frequency of 1280 Hz is travelling at 120 km/h. (problem 65) What frequencies does an observer standing next to the road hear as the car approaches and as it recedes What frequencies are heard in a car travelling at 90 km/h in the opposite direction before and after passing the police car? The police car passes a car traveling in the same direction at 80 km/h. What two frequencies are heard in the car?

Explanation / Answer

2)   First, find the intensity of the 120dB
B = 10 log (I/Io)
120dB = 10 log (I/10^-12)
12dB = log (I/10^-12)
10^12 = I/10^-12
10^0 = I
1W/m^2 = I

Now, calculate the intensity of 20dB.
B = 10 log (I/Io)
20dB = 10 log (I/10^-12)
2 = log (I/10^-12)
10^2 = I/10^-12
10^-10W/m^2 = I

So, the intensity of sound at pain level (120dB) is 10^10 times that of a whisper at 20dB

3) a comparison could be:
=10 log (P2÷P1)
=10 log (150 ÷ 100)
= 1.7609
So the 150 W is likely to be 1.7609 times more powerful

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