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

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be 'n' loops and $\lambda$ be the wavelength.
∴ $\lambda$ = $2\times \text{S}\mathrm{eparation}\mathrm{between}\mathrm{the}\mathrm{consecutive}\mathrm{nodes}$


${\lambda }_1=2\times 2=4\mathrm{cm}$


${\lambda }_2=2\times 1.6=3.2\mathrm{cm}$
Length of the wire is L.

In the first case:
$L=\left(\frac{n{\lambda }_1}{2}\right)$
In the second case:
$$\begin{gathered} \mathrm{L}=\left(n+1\right)\frac{{\lambda }_2}{2} \\ \Rightarrow \frac{n{\lambda }_1}{2}=\left(n+1\right)\frac{{\lambda }_2}{2} \\ \Rightarrow n\times 4=\left(n+1\right)\left(3.2\right) \\ \Rightarrow 4n-3.2n=3.2 \\ \Rightarrow 0.8n=3.2 \\ \Rightarrow n=4 \end{gathered}$$
∴ Length of the string, $L=\frac{\left(n{\lambda }_1\right)}{2}=\frac{\left(4\times 4\right)}{2}=8\mathrm{cm}$