A string fixed at both ends vibrates in a res | vedclass

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(B) Let the string have length $L$. The distance between consecutive nodes is $\lambda / 2$.In the first case,$\lambda_1 / 2 = 2.0 \, cm$,so $\lambda_

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$8$

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(B) Let the string have length $L$. The distance between consecutive nodes is $\lambda / 2$.In the first case,$\lambda_1 / 2 = 2.0 \, cm$,so $\lambda_1 = 4.0 \, cm$.The length of the string is $L = n(\lambda_1 / 2) = n(2.0) = 2n$,where $n$ is the number of loops.In the next higher resonant frequency,the number of loops becomes $(n+1)$.The new separation between nodes is $\lambda_2 / 2 = 1.6 \, cm$,so $\lambda_2 = 3.2 \, cm$.The length of the string is $L = (n+1)(\lambda_2 / 2) = (n+1)(1.6)$.Since the length $L$ is constant,we equate the two expressions:$2n = 1.6(n+1)$$2n = 1.6n + 1.6$$0.4n = 1.6$$n = 4$.Substituting $n$ back into the expression for $L$:$L = 2(4) = 8 \, cm$.

$A$ string fixed at both ends vibrates in a resonant mode with a separation of $2.0 \, cm$ between consecutive nodes. For the next higher resonant frequency,this separation is reduced to $1.6 \, cm$. The length of the string is .... $cm$.

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