4) a bat hanging from the wall of a cave emits ultrasound waves at a frequency o

Question

4) a bat hanging from the wall of a cave emits ultrasound waves at a frequency of 50,000Hz. A moth flies toward the bat 13m/s. what frequency does the moth hear? if now the bat flies directly toward the moth at 20 m/s, what frequency does the moth hear? what frequency of its reflected signal would the flying bat hear off the approaching moth? b) sound from two sources has decibel levels of 53 dB and 35 dB at one point in space. what is the ratio of the sound intensities at this point? if both sources of sound are equally powerful what is the ratio of distance to the two sources?

Explanation / Answer

Using Doppler eqn we have fL = fS*(v + vL)/(v+vS) vL = 0 and vS = -13 So vL (frequency moth hears) = 50.0kHZ*343/(343-13) = in kHz ( calculate)............................. b) If the speed is increased the frequency will increase also. Doppler effect is that as the source approaches the listener the listener hears a higher pitch or frequency .........................The decibel scale is defined as beta = 10*log_10(I/I0)...... Apply subscripts 1 and 2 to correspond to the conditions given: beta1 = 10*log_10(I1/I0)............................ beta2 = 10*log_10(I2/I0).................................. Subtract (you will see why), beta2 - beta1: beta2 - beta1 = 10*(log_10(I2/I0) - log_10(I1/I0))...... Combining the logarithm, and you get: beta2 - beta1 = 10*log_10(I2/I1)....... Solve for the ratio of I's: (beta2 - beta1)/10 = log_10(I2/I1) ................... Exponentiate both sides with base 10: 10^((beta2 - beta1)/10) = I2/I1.............. Thus: I2/I1 = 10^((beta2 - beta1)/10)...................Data: beta2:=53 dB; beta1:=35 dB; Result: I2/I1 =calculate...............................

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