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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

2dn

c

2d(n+1)

d

d(n+1)

answer is A.

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Detailed Solution

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nth overtone = n+1
harmonic ⇒ length= (n+1)λ2 
but d=λ4  
 

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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