Question

A string is fixed at both ends vibrates in a resonant mode with a separation $2.0cm$ between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to $1.6cm$. The length of the string is

• A. $4.0cm$
• B. $8.0cm$
• C. $12.0cm$
• D. $16.0cm$

Answer 1

The correct option is B $8.0cm$

Let $L$ be the length of the string, $\lambda$ be the distance between two consecutive nodes, $n$ be the frequency.
Therefore,
$L=n\lambda ...........................\left(1\right)$
Also, for next higher frequency i.e. $(n+1)$ the distance between two consecutive nodes is ${\lambda }^{\prime }$(say), then
$L=(n+1){\lambda }^{\prime }.......................\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$, we get,
$(n+1){\lambda }^{\prime }=n\lambda$
${\lambda }^{\prime }=n\lambda -n{\lambda }^{\prime }$
${\lambda }^{\prime }=n(\lambda -{\lambda }^{\prime })$
$n=\frac{1.6}{2-1.6}=\frac{1.6}{0.4}=4$
$\Rightarrow L=4\times 2=8cm$.................from$\left(1\right)$