A string is vibrating in its fifth overtone b | vedclass

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(B) The length of the string is $L = 2.4 ~m$.In the $n^{th}$ overtone,the string vibrates in $(n+1)$ loops.For the fifth overtone,$n = 5$,so the numbe

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$0.2$

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(B) The length of the string is $L = 2.4 ~m$.In the $n^{th}$ overtone,the string vibrates in $(n+1)$ loops.For the fifth overtone,$n = 5$,so the number of loops is $5 + 1 = 6$.The length of each loop is the distance between two successive nodes,which is $\lambda/2 = L / (n+1) = 2.4 / 6 = 0.4 ~m$.The distance between a node and the adjacent antinode is always $\lambda/4$.Since the distance between two successive nodes is $\lambda/2 = 0.4 ~m$,the distance between a node and an antinode is $\frac{0.4 ~m}{2} = 0.2 ~m$.

$A$ string is vibrating in its fifth overtone between two rigid supports $2.4 ~m$ apart. The distance between successive node and antinode is (in $~m$)

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