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Q.

A wire length L and mass per unit length 6.0×103kgm1 is put under tension of  540N. Two consecutive frequencies that it resonates at are: 420Hz and 490Hz . Then L in meters is

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a

1.1m

b

2.1m

c

5.1m

d

8.1m

answer is B.

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Detailed Solution

The difference in the frequencies of two successive harmonics is equal to fundamental frequency      n=490420=70Hz
n=12LTμ       70=12×L5406×103   L=1140×300=2.142 m 

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