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A string of length $1\,\,m$ and linear mass density $0.01\,\,kgm^{-1}$ is stretched to a tension of $100\,\,N.$ When both ends of the string are fixed, the three lowest frequencies for standing wave are $f_1, f_2$ and $f_3$. When only one end of the string is fixed, the three lowest frequencies for standing wave are $n_1, n_2$ and $n_3$. Then
$n_3 = 5n_1 = f_3 = 125 \,\,Hz$
$f_3 = 5f_1 = n_2 = 125 \,\,Hz$
$f_3 = n_2 = 3f_1 = 150 \,\,Hz$
$n_2 =\frac{{{f_1} + {f_2}}}{2} = 75 \,\,Hz $
Solution
When both ends are fixed, the string forms a length half the wavelength. That is, it has two nodes at the ends. For the next frequency, it will have the length equals the wavelength. So, the general formula for length of the string becomes $L=n \lambda / 2$
For the string fixed on only one end, there is always an anti node at one end and a node at the other end. So, the length of the string gets divided into $1 / 4 t h$ of the wavelength $(\lambda) .$ The general formula for the length of the string is $L^{\prime}=n \lambda / 4$
The frequency $f$ becomes $V / \lambda, V$ is the velocity. In the first case, frequency $f=n V / 2 L$ $n=1,2,3, \dots$
In the second case, it is $n V / 4 L, \quad n=1,3,5,7 \ldots \ldots$ because of the length of the string will always have a half wave present. This makes $n$ an odd number.
For the first case$:$
$f_{1}=1 / 2 L(\sqrt{(T / \mu)})=50 H z=V / 2 \times L$
$f_{2}=2 \times f_{1}=100 H z=V / L$
$f 3=3 \times f_{1}=150 H z=3 V / 2 \times L$
Second case$:$
$n_{1}=V / 4 L$
$n_{2}=3 V / 4 L=\left(f-1+f_{2}\right) / 2=(100+50) / 2=75 H z$
