Question

A 75cm string fixed at both ends produces resonant frequencies of 384Hz and 288Hz without there being any other resonant frequency between these two. Wave speed for the string is:
A. $144m/s$
B. $216m/s$
C. $108m/s$
D. $72m/s$

Answer 1

Hint: When string is fixed from both ends nodes will be formed at both ends. When a string is only fixed from one end and set free at the other end node is formed at fixed end and antinode is formed at the free end. Wavelengths can be found as length of string(l) is given and velocity can be found as consecutive resonant frequencies are also given.

Formula used:
$v = f\lambda $
$f = \dfrac{{nv}}{{2l}}$

Complete answer:
Let the speed of wave in the string be ‘v’ and length of the string be ‘l’ and frequency be ‘f’. In the resonating forms as the wavelength changes for every harmonic, frequency also changes. We have the equation $v = f\lambda $
When string is fixed at both ends initially one loop will be formed and hence wavelength of that loop be ${\lambda _1}$
For one loop distance is $\dfrac{{{\lambda _1}}}{2}$
So initially $\dfrac{{{\lambda _1}}}{2} = l$, ${\lambda _1} = 2l$
$ \Rightarrow v = f\lambda $
$ \Rightarrow {f_1} = \dfrac{v}{{2l}}$ … eq1
Then next it forms two loops this time wave length be ${\lambda _2}$
So distance would be ${\lambda _2} = l$
$v = f\lambda $
$ \Rightarrow {f_2} = \dfrac{{2v}}{{2l}}$ … eq2
Then it forms three loops now wavelength is ${\lambda _3}$
So distance would be $\dfrac{{3{\lambda _3}}}{2} = l$
$v = f\lambda $
$ \Rightarrow {f_3} = \dfrac{{3v}}{{2l}}$ … eq3
From equation 1 and 2 and 3
We can form a general expression of the resonating frequencies. That expression will be
$f = \dfrac{{nv}}{{2l}}$
Where ‘f’ is the resonating frequency and ‘n’ is the positive integer.
If we clearly observe, if we subtract any resonant frequency from the previous resonant frequency we get the fundamental frequency $\dfrac{v}{{2l}}$
$ \Rightarrow {f_2} - {f_1} = {f_3} - {f_2} = {f_4} - {f_3} = {f_{n + 1}} - {f_n} = \dfrac{v}{{2l}}$
This literally means that the magnitude of difference between two consecutive frequencies will be fundamental frequency.
So
$\eqalign{
  & 384 - 288 = \dfrac{v}{{2l}} \cr
  & \Rightarrow \dfrac{v}{{2l}} = 96 \cr
  & \Rightarrow v = 96 \times 2 \times \left( {\dfrac{{75}}{{100}}m} \right) \cr
  & \Rightarrow v = 144m/s \cr} $

Hence option A will be the answer.

Note:
In case of string fixed at both ends or fixed at one end length of the string is constant but wavelengths are varying and hence frequencies vary. If one clearly observes two ends, the fixed case ratio of frequencies will be 1:2:3:4:5….. and in one end fixed case it would be 1:3:5:7…. A and so on. So without calculations if we find first fundamental frequencies in both cases we can find the other frequencies.