A string is vibrating in its fifth overtone b | vedclass

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(B) For a string fixed at both ends,the $n^{th}$ overtone corresponds to the $(n+1)^{th}$ harmonic.Here,the fifth overtone is the sixth harmonic $(n=6

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$0.2$

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(B) For a string fixed at both ends,the $n^{th}$ overtone corresponds to the $(n+1)^{th}$ harmonic.Here,the fifth overtone is the sixth harmonic $(n=6)$.The length of the string $L = 2.4 \ m$.The condition for the $n^{th}$ harmonic is $L = n \frac{\lambda}{2}$.Substituting the values: $2.4 = 6 	imes \frac{\lambda}{2}$.This gives $\frac{\lambda}{2} = \frac{2.4}{6} = 0.4 \ m$.Thus,$\lambda = 0.8 \ m$.The distance between a node and the successive antinode is always $\frac{\lambda}{4}$.Therefore,distance $= \frac{0.8 \ m}{4} = 0.2 \ m$.

$A$ 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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