A ukulele string is stretched to a length of 0.49 m and both ends are fixed. A)

Question

A ukulele string is stretched to a length of 0.49 m and both ends are fixed.

A) If the density of the string is 0.0015 kg/m and its tension is 76.1 N, what is the funda- mental frequency?

Answer in units of Hz

B) Two ukulele strings, of equal length and mass are tuned so that the first string, when it vi- brates at twice its fundamental frequency, has the same frequency as the second string when it vibrates at three times its fundamental fre- quency. The tension of the first string is 31 N.

Calculate the tension F2 of the second string.

Answer in units of N

Explanation / Answer

A) fo = (1/2L)*sqrt(T/u)

fo = (1/9.8)*sqrt(76.1/0.0015)= 22.98 Hz

B) 2fo = 3fo'

2*(1/2L)*sqrt(T1/u) = 3*(1/2L)*sqrt(T2/u)

4T1 = 9T2

T2 = (4*31)/9 = 13.78 N

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