Question

A string is stretched between fixed points separated by $75cm$. It is observed to have resonant frequencies of $420Hz$ and $315Hz$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is
(a) $10.5Hz$
(b) $105Hz$
(c) $1.05Hz$
(d) $1050Hz$

Answer 1

We can approach this problem by solving the equation of consecutive frequencies at $nth$ and $\left( {n + 1} \right)$ th harmonics. We know that ${f_n} = \dfrac{{nv}}{{2L}}$ and ${f_{n + 1}} = \dfrac{{(n + 1)v}}{{2L}}$ are the consecutive frequencies . Here $L$ is length of the string and $v$ is speed of the wave.

Complete step by step answer:
For strings fixed at both the ends, resonant frequencies are given by the equation $f = \dfrac{{nv}}{{2L}}$, where symbols have their usual meanings. it is given that $315Hz$ and $420Hz$ are two consecutive resonant frequencies, let these be $nth$ and $\left( {n + 1} \right)$ th harmonics respectively so we can put the values of these frequencies in the above equation of consecutive frequencies and get the following equations;
$\begin{gathered} 315 = \dfrac{{nv}}{{2L}}.........(1) \\\ 420 = \dfrac{{\left( {N + 1} \right)v}}{{2L}}..........(2) \\\ \end{gathered}$

By dividing both equations 1 and 2 we get the following equation and after solving this equation we can get the value of $n$.
$\begin{gathered} \dfrac{{315}}{{420}} = \dfrac{n}{{n + 1}} \\\ or \\\ n = 3 \\\ \end{gathered}$
Thus the lowest resonant frequency is $\dfrac{{315}}{n} = 105Hz$. Hence when a string is stretched between fixed points separated by $75cm$. It is observed to have resonant frequencies of $420Hz$ and $315Hz$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is $105Hz$.

So option (b) is the correct answer to this problem.

Note:
We have calculated the lowest frequency in this problem. We have used the formula of consecutive resonant frequency in order to calculate the value of harmonics. When we divide the value of harmonic by resonant frequency we get the value of lowest resonant frequency. Some important information is given in the question which is helpful in finding the answer such as a string is stretched between fixed points separated by $75cm$. It is observed to have resonant frequencies of $420Hz$ and $315Hz$.