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

Let there be 'n' loops in the 1st case

$\Rightarrow$ Length of the wire,

$L=(\frac{n{\lambda }_1}{2}$

$({\lambda }_1=2\times 2=4cm)$

$\Rightarrow$ Length of the wire,

$L=\left\{\right(n+1\left)\frac{{\lambda }_2}{2}\right\}$

$({\lambda }_2=2\times (1.6)=3.2cm)$

$\Rightarrow \frac{n{\lambda }_1}{2}=(n+1)\frac{{\lambda }_2}{2}$

$\Rightarrow n\times 4=(n+1)\left(3.2\right)$

$\Rightarrow 4n-\left(3.2\right)n=3.2$

$\Rightarrow 0.8n=3.2$

$\Rightarrow n=4$

$\therefore$ Length of the string,

$L=\frac{\left(n{\lambda }_1\right)}{2}=\frac{(4\times 4)}{2}=8cm$