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 1

Step1: Given data and assumptions.

Separation distance between first node, $\lambda =2.0cm$

Separation distance between next node, $\lambda \text{'}=1.6cm$

Let the frequency be $n$.

And the length of the string be $L$.

Step 2: Formula used:

$L=n\lambda$

Where $L$is length, $n$ is the frequency, $\lambda$ is the separation distance.

Step 3: Finding the length of the string.

The frequency, $n$ of the first node, $\lambda$.

$L=n\lambda$ ……$\left(i\right)$

Again,

The frequency, $n+1$ of the next node, $\lambda \text{'}$.

$L=\left(n+1\right)\lambda \text{'}$ ……..$\left(ii\right)$

From equation $\left(i\right)$ and $\left(ii\right)$ we get,

$n\lambda =\left(n+1\right)\lambda \text{'}$

$\Rightarrow$ $n\times 2=\left(n+1\right)\times 1.6$

$\Rightarrow$ $2n=1.6n+1.6$

$\Rightarrow$$2n-1.6n=1.6$

$\Rightarrow$ $n=\frac{1.6}{0.4}$

$\Rightarrow$ $n=4$ ……$\left(iii\right)$

Now, substitute equation $\left(iii\right)$ in $\left(i\right)$ we get,

$L=4\times 2$

$\Rightarrow$$L=8cm$

Hence, the length of the string is $\mathbf{8}\mathbf{}\mathit{c}\mathit{m}$.