Obtain the equation of frequency of oscillations in a string tied at both ends.

(N/A) Consider a string of length $L$ fixed at both ends under tension $T$. Let $\mu$ be the linear mass density of the string.
The wave equation for a stationary wave is given by $y(x, t) = A \sin(kx) \cos(\omega t)$.
Since the string is fixed at $x = 0$ and $x = L$,the boundary conditions are $y(0, t) = 0$ and $y(L, t) = 0$.
Applying the boundary condition at $x = L$,we get $\sin(kL) = 0$,which implies $kL = n\pi$ for $n = 1, 2, 3, \dots$.
Since $k = \frac{2\pi}{\lambda}$,we have $\frac{2\pi L}{\lambda} = n\pi$,which gives $\lambda = \frac{2L}{n}$.
The speed of the wave in the string is $v = \sqrt{\frac{T}{\mu}}$.
Using the relation $v = f\lambda$,where $f$ is the frequency,we get $f = \frac{v}{\lambda}$.
Substituting $\lambda = \frac{2L}{n}$,we obtain $f = \frac{nv}{2L} = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$,where $n = 1, 2, 3, \dots$ represents the mode of vibration.