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A string of mass per unit length mu = 6 ...

A string of mass per unit length `mu = 6 × 10^(-3)` kg/m is fixed at both ends under the tension 540 N. If the string is in resonance with consecutive frequencies 420 Hz and 490 Hz. Then find the length of the string?

A

8.1 m

B

2.1 m

C

1.1 m

D

5.1 m

Text Solution

Verified by Experts

The correct Answer is:
B
Fundamental frequency = 490 -420 = 70 Hz
`70=1/(2l)sqrt((T)/(mu))implies70 =1/(2l)sqrt((540)/(6xx10^(-3)))`
`implies l = (1)/(2xx70)sqrt(90 xx10^(3))=300/140`
`l~~2.14 m`
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