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.
