A string with a mass density mu = 1.30 Times l0-3 kg / m is under a tension of F

Question

A string with a mass density mu = 1.30 Times l0-3 kg / m is under a tension of F = 356 N and is fixed at both ends One of its resonance frequencies is 148.0 Hz. The next higher resonance frequency is 185.0 Hz. What is the fundamental frequency of this string ? Which harmonic does the resonance frequency at 148.0 Hz correspond to? (i.e. what is n at this frequency ?) What is the length of the string ? Now, suppose the same string is detached at one end and connected by a ring to a frictionless post, so that it can move freely. Find the wavelength of the first (fundamental) harmonic. What is the frequency of the third (n = 3) harmonic in this case ? Tries 0 / 99

Explanation / Answer

gieven

nV/2L =148----1 and

(n+1)V/2L =185-----2

2/1

n+1/n =185/148

148n+148 =185n

37n=148

n=4

a)

f1 =148/4 =37 Hz

b)

n=4

c)

V=sqrt(T/u) =sqrt(356/1.3*10^-3)

V=523.3 m/s

V/2L=37

523.3/2L=37

L=7.07 m

d)

lambda =4*L=4*7.07 =28.28 m

e)

f1=V/lambda =523.3/28.28

f1=18.5 Hz

f3=3f1 =55.5 Hz

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