Question
A string, fixed at both ends, vibrates in a resonant mode with a separation of 2.0cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6cm. Find the length of the string.
✓
Answer
Let there be ‘n’ loops in the 1st case ⇒ length of the wire, $\text{l}=\Big(\frac{\text{n}\lambda_1}{2}\Big)$ $[\lambda_1=2\times2=4\text{cm}]$ So there are (n + 1) loops with the 2nd case 
⇒ length of the wire, $\text{l}=\Big\{\frac{(\text{n}+1)\lambda_2}{2}\Big\}$ $[\lambda=2\times1.6=3.2\text{cm}]$ $\Rightarrow\frac{\text{n}\lambda_1}{2}=\frac{(\text{n}+1)\lambda_2}{2}$
$\Rightarrow\text{n}\times4=(\text{n}+1)(3.2)$
$\Rightarrow\text{n}=4$
$\therefore\ $length of the string, $\text{l}=\frac{\text{n}\lambda_1}{2}=8\text{cm}.$
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