Question

A string, fixed at both ends, vibrates in a resonant mode with a separation of 2⋅0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1⋅6 cm. Find the length of the string.

Answer 1

Length of String is 8 cm

Explanation:

Let there be ‘n’ loops in the 1st case-

Length of wire, l = $\frac {(n\lambda_1)}{2}$ [$\lambda = 2 \times2 = 4$ cm]

So there are (n+ 1) loops with the II case

length of the wire,$l = {\frac {(n + 1)\lambda_2}{2}\       [\lambda = 2 \times 1.6 = 3.2 cm]$

$\frac {n\lambda_1}{2} = \frac {(n + 1)\lambda_2}{2}$

$n \times 4$ = (n + 1)(3.2)

n = 4

So, length of the string, l = $\frac {n\lambda_1}{2}$ = 8 cm