A stretched string fixed at each end has a mass of 49.0 g and a length of 8.20 m

Question

A stretched string fixed at each end has a mass of 49.0 g and a length of 8.20 m. The tension in the string is 48.0 N. (a) Determine the positions of the nodes and antinodes for the third harmonic. (Enter your answers from smallest to largest distance from one end of the string.) nodes: 0 2.37 Your response differs from the correct answer by more than 10%. Double check your calculations. m 5.47 Your response differs from the correct answer by more than 10%. Double check your calculations. m antinodes: 1.18 Your response differs from the correct answer by more than 10%. Double check your calculations. m 3.54 Your response differs from the correct answer by more than 10%. Double check your calculations. m 5.9 Your response differs from the correct answer by more than 10%. Double check your calculations. m (b) What is the vibration frequency for this harmonic? The human ear canal is about 2.3 cm long. If it is regarded as a tube open at one end and closed at the eardrum, what is the fundamental frequency around which we would expect hearing to be most sensitive?

Explanation / Answer

(1) ans

Given that

mass m=0.049 kg

length of the string L=8.2 m

tension of the string T=48 N

now we find the nodes and antinodes for third hormonic

nodes N1=8.2 m

node N2=3*8.2=24.6 m

node N3=4*8.2=32.8 m

antinodes A1=8.2/2=4.1 m

antinodes A2=2*8.2/2=8.2 m

antinodes A3=3*8.2/2=12.3 m

now we find the fundamental frequency for this hormonic

fundamental frequency f=1/2L{T/(m/l)}^1/2

=1/2*8.2{48/(0.049/8.2)}^1/2

=5.5 Hz

the frequency for third harmonic f3=3f=3*5.5=16.5 Hz

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