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
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For a taut string between two ends$:$
$n^{\text {th}}$overtone$=(n+1)$harmonic
$n^{\text {th}}$overtone$=(n+1)$harmonic
$\frac{(n+1)}{2} \lambda=L \ldots \ldots \ldots .(L=\text {lengtho $f$ string}) \ldots \ldots \ldots .(1)$
Distance between node and antinode is given by $\frac{\lambda}{4}$ which is equal to $”d”$ according to
the question. $\left(\frac{\lambda}{4}=d o r \lambda=4 d\right)$
Substituting the value of $\mathrm{d}$ in equation $(1)$
$\frac{(n+1) 4 d}{2}=L$
$(n+1) 2 d=L$
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