Question

A string of length 1m and linear mass density 0.01kgm$^{−1}$ is stretched to a tension of 100N. When both ends of the string are fixed, the three lowest frequencies for standing waves are $f_1$​, $f_2$​ and $f_3$​. When only one end of the string is fixed, the three lowest frequencies for standing waves are $n_1$​, $n_2$​ and $n_3$​. Then
\eqalign{ & A.{n_3} = 5{n_1} = {f_3} = 125Hz \cr & B.{f_3} = 5{f_1} = {n_2} = 125Hz \cr & C.{f_3} = {n_2} = 3{f_1} = 150Hz \cr & D.{n_2} = \dfrac{{{f_1} + {f_2}}}{2} = 75Hz \cr}

Answer 1

when string is fixed from both ends noses 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. Since tension and linear mass density are given it is possible to find out the velocity of the standing wave in the string and wavelengths can be found as length of string(l) is given and finally frequencies can be found.

Formula used:
$v = f\lambda$
$v = \sqrt {\dfrac{T}{\mu }}$

Complete answer:
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 $\dfrac{{{\lambda _1}}}{2} = l$, ${\lambda _1} = 2l = 2$ … eq1
Then next it forms two loops this time wave length be ${\lambda _2}$
So distance would be ${\lambda _2} = l$ ${\lambda _2} = 1$ … eq2

Then it forms three loops now wavelength is ${\lambda _3}$
So distance would be $\dfrac{{3{\lambda _3}}}{2} = l$ ${\lambda _3} = \dfrac{2}{3}$ … eq3
Velocity of wave would be $\sqrt {\dfrac{T}{\mu }}$ where T is tension and $\mu$ is linear mass density
Velocity(v) = $\sqrt {\dfrac{{100}}{{0.01}}} = 100m/s$
$v = f\lambda$
f is frequency
From equation 1 and 2 and 3
We get
${f_1} = \dfrac{v}{{{\lambda _1}}} = \dfrac{{100}}{2} = 50Hz$
${f_2} = \dfrac{v}{{{\lambda _2}}} = \dfrac{{100}}{1} = 100Hz$
${f_3} = \dfrac{v}{{{\lambda _3}}} = \dfrac{{100}}{2} \times 3 = 150Hz$
now when string is free at one side fundamental frequency would be ${n_1} = \dfrac{v}{{4l}} = \dfrac{{100}}{4} = 25Hz$
And the ratio of ${n_1},{n_2},{n_3}$ would be $1:3:5$
So ${n_2} = 3 \times 25 = 75,{n_3} = 5 \times 25 = 125$
So ${n_2} = \dfrac{{{f_1} + {f_2}}}{2} = 75Hz$

Hence option D would be correct.

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 in the first case i.e both ends, the fixed case ratio of frequencies will be 1:2:3:4:5….. and in the second case it would be 1:3:5:7…. and so on. So without calculations if we find first fundamental frequencies in both cases we can find the other frequencies.