Question

A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string

Answer 1

Step 1:Given:

$$\begin{gathered} f_1=420\text{Hz} \\ {f}_2=315\text{Hz} \end{gathered}$$

$l=75\text{cm}$

Step 2:To Find:

lowest resonant frequency

Step 3:Calculate Lowest Resonant frequency:

$\frac{nv}{2l}=315\text{Hz}$ (1)

$\frac{\left(n+1\right)v}{2l}=420\text{Hz}$ (2)

After solving equation (1) and (2) get the value of $n$ is $3$.

$$\begin{gathered} f_{lowest}=\frac{f_2}{n} \\ =\frac{315\text{Hz}}{3} \\ =105\text{Hz} \end{gathered}$$