Part A Estimate the frequency of the \"sound of the ocean\" when you put your ea

Question

Part A

Estimate the frequency of the "sound of the ocean" when you put your ear very near a 18-

cm-diameter sea-shell. The speed of sound in air is 343 m/s.

Express your answer to two significant figures and include the appropriate units.

Part B

At T = 14 C, how long must an open organ pipe be to have a fundamental frequency of 392 Hz ? The speed of sound in air is v(331+0.60T)m/s, where T is the temperature in C.

Express your answer to three significant figures and include the appropriate units.

Part C

If this pipe is filled with helium at 20C and 1 atm, what is its fundamental frequency? The speed of sound in helium is 1005 m/s.

Express your answer to three significant figures and include the appropriate units.

Part D

A guitar string produces 3 beats/s when sounded with a 343-Hz tuning fork and 8 beats/s when sounded with a 348-Hz tuning fork.What is the vibrational frequency of the string?

Part E

Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 15 m/s , the driver at rest hears a beat frequency of 9.0 Hz .

What is the frequency the horns emit? Assume T=20C.

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Part A

assuming the seashell is closed at both ends, for funamental

lambda/4 = 0.18 or lambda =0.72 m

hence the required frequency is 343/0.72 = 476.39 Hz or its multiple.

Part B:

c=339.4 m/s at 21C
open pipe resonates when length is 1/2 wave length
c=f

L=/2=c/(2f) = 339.4/(2*392) =0.4329 m

Part C:

f = v/(2L) = 1005/(2*0.4329)=1160.77 Hz

Part D:

In each case the quitar is producing the difference frequency between the tuning fork and the vibrational frequency of the string.

case 1: 343 - 340 = 3 beats/sec

case 2: 348 - 340 =8 beats/sec

vibrational frequency of the string = 340 Hz

Part E:

  Due to Doppler effect the frequency observed from the moving car f' is greater than from the car at rest.

The beat frequency arising from the interference of the two sound waves equals the difference of frequencies. i.e.
f_beat = f' - f

The observed frequency is given by
f' = fc /(c - v)
where
v the speed of the car toward the observer
c the speed of sound
at T=20°C c = 343.2m/s

hence:
f_beat = fc/(c - v) - f
<=> f_beat = fv/(c - v)
<=> f_beat = f/((c/v) - 1)

Solve for f
=> f = f_beat ((c/v) - 1)
= 9.0Hz ( (343.2m/s/15m/s) - 1 )
= 196.92 Hz

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