A ukulele string is stretched to a length of 0.6 m and both ends are fixed. If t

Question

A ukulele string is stretched to a length of 0.6 m and both ends are fixed. If the density of the string is 0.0018 kg/m, and its tension is 65.1 N, what is the fundamental frequency? Answer in units of Hz. Two ukulele strings, of equal length and mass are tuned so that the first string, when it vibrates at twice its fundamental frequency, has the same frequency as the second string when it vibrates at three times its fundamental frequency. The tension of the first string is 52 N Calculate the tension F_2 of the second string. Answer in units of N.

Explanation / Answer

fundamental frequency f = (1/2L)*sqrt(T/u)

T = 65.1 N

linear density u = 0.0018 kg/m

L = 0.6 m

f = (1/(2*0.6))*sqrt(65.1/0.0018)

f = 158.5 hz

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Q4

fundamental frequency of strig 1 fo1 = (1/2L)*sqrt(F1/u)

fundamental frequency of strig 2 fo2 = (1/2L)*sqrt(F2/u)

given

2*fo1 = 3*fo2

2*(1/2L)*sqrt(F1/u) = 3*(1/2L)*sqrt(F2/u)

2*sqrtF1 = 3*sqrtF2

2*sqrt52 = 3*sqrtF2

F2 = 23.1 N <<<<---answer

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