Q.

If n1, n2 and n3 are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency n of the string is given by

Moderate

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1n=1n1+1n2+1n3

n=n1+n2+n3

answer is 1.

(Detailed Solution Below)

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Detailed Solution

n1=12l1Tμ . . . . (i)n2=12l2Tμ . . . . .(ii)n3=12l3Tμ . . . . . (iii) andn=12lTμ . . . . . (iv) From eqn. (iv), we get 1n=2lTμ As l=l1+l2+l3∴ 1n=2l1+l2+l3Tμ=2l1Tμ+2l2Tμ+2l3Tμ=1n1+1n2+1n3 (using (i), (ii) and (iii))

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