Question

A string, 0.28 m long and vibrating in its third harmonic, excites an open pipe that is 0.82 m long into its second overtone resonance. The speed of sound in air is 345 m/s. The speed of transverse waves on the string is closest to:__________
a. 110
b. 98
c. 100
d. 120
e. 91

Answer 1

Answer: 98.

Explanation: it has been stated in the question that sound wave in the string and in the pipe resonated at a specific frequency, this simply implies that the frequency of sound wave in the string equals frequency of sound wave in pipe.

Fs = Fp.

The length (l) of the string is 0.28m and it is vibrating at it third harmonic.

The length of stationary wave on a string at third harmonic is given below as

l = 3λ/2

Where λ = wavelength of sound wave in pipe (λs)

By substituting l = 0.28m into the equation above, we have that

0. 28 = 3λs/2

3λs = 0.28 * 2

3λs = 0.56, λs= 0.56/ 3

λs = 0.187m

Thus the wavelength of wave in the string is 0.187m.

Sound from the string in the pipe is produced at the second overtone ( which is the third harmonic).

Therefore the length of air in the pipe at second overtone ( third harmonic) is given below as

l = 5λp/ 4, we need to get the wavelength of sound in the pipe.

Thus

λp = 4*l/5

λp = 4 * 0.82 / 5

λp = 0.656m.

The velocity of sound waves produced in the pipe is 345m/s thus the frequency of sound in the pipe is gotten using the formulae below

V = fpλp

V= velocity of sound in pipe, fp = frequency of sound in pipe, λp= wavelength of sound in pipe

345 = f / 0.656

fp = 525.92Hz.

As stated in the question, the frequency of sound in pipe is the same as that in string (fp = fs = f) , thus to get the velocity of sound wave in string we use the same formulae of

v = fλ

Where f = frequency of sound in pipe = frequency of sound in string = 525.92Hz.

λ = wavelength of sound in string = 0.187m

Thus v = 525.92 * 0.187 = 98.34 which is closest to 98.