Question

A taut string at both ends vibrates in its $n^{th}$ overtone. The distance between adjacent Node and Antinode is found to be '$d$'. If the length of the string is $L$, then

• A. $-L=2d(n+1)$
• B. $L=d(n+1)$
• C. $L=2dn$
• D. $L=2d(n-1)$

Answer 1

Answer: (A)

$-L=2d(n+1)$

For a taut string between two ends:
$n^{th}overtone=(n+1)harmonic$
$n^{th}overtone=(n+1)harmonic\frac{(n+1)}{2}\lambda =L..............(L=lengthofstring).............\left(1\right)$

Distance between node and antinode is given by $\frac{\lambda }{4}$ which is equal to $"d"$ according to the question. $(\frac{\lambda }{4}=dor\lambda =4d)$

Substituting the value of d in equation (1)

$\frac{(n+1)4d}{2}=L(n+1)2d=L$
Hence option A is correct