A string is fixed at both ends vibrates in a resonant mode with a separation $2.0 \,\,cm$ between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to $1.6\,\, cm$. The length of the string is .... $cm$
Diffcult
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Let there be $n$ loops in the $1\, st$ case
$\rightarrow$ Length of the wire$L$ $=\left(\left(\mathrm{n} \lambda_{1}\right) / 2\right)$
$\left(\lambda_{1}=2 \times 2=4 \mathrm{cm}\right)$
$\rightarrow$ Length of the wire $L_{l}=\left\{(n+1) \frac{\lambda_{2}}{2}\right\}$
$\left(\lambda_{2}=2 \times(1.6)=3.2 \mathrm{cm}\right)$
$\rightarrow \frac{n \lambda_{1}}{2}=(n+1) \frac{\lambda_{2}}{2}$
$\rightarrow n \times 4=(n+1)(3.2)$
$\rightarrow 4 n-(3.2) n=3.2$
$\rightarrow 0.8 n=3.2$
$\rightarrow n=4$
Length of the string
$L=\frac{n \lambda_{1}}{2}=\frac{(4 \times 4)}{2}=8 \mathrm{cm}$
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