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Q.

A taut string fixed at both ends vibrates in nth overtone. The separation between adjacent node and antinode is equal to ‘d’. Then string length is

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a

2d(n+1)   

b

 d(n+1)

c

2dn

d

2d(n–1)

answer is A.

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Detailed Solution

⇒d=λ4

∴λ=4d

⇒l=(n+1)λ2

⇒l=(n+1)4d2

∴l=2d(n+1)

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