A string of length L secured at both ends vibrates in its...

The general equation of a standing wave is $y(x, t) = f(x) g(t)$.

Since the string is fixed at $x = 0$ and $x = L$, the displacement at these points must be zero. This requires the spatial part to be a sine function: $f(x) = A \sin(kx)$.

For the fundamental mode, the string vibrates in a single loop, which means $L = \dfrac{\lambda}{2} \Rightarrow \lambda = 2L$.

The wave number is $k = \dfrac{2\pi}{\lambda} = \dfrac{2\pi}{2L} = \dfrac{\pi}{L}$.

Thus, the spatial part is $f(x) = A \sin$\left(\dfrac{\pi x}{L}\right).

The temporal part is given by $g(t) = \sin(\omega t + \phi)$.

At $t = 0$, the middle point of the string ($x =$\dfrac{L}{2}) crosses the mean position ($y = 0$), which gives $g(0) = 0$. This implies $\phi = 0$ or $\pi$.

Since the particle is travelling towards the positive $y$-direction, the velocity \dfrac{\partial y}{\partial t} at $t = 0$ must be positive.

The velocity is proportional to $\omega \cos(\omega t + \phi)$. For this to be positive at $t = 0$, we must have $\phi = 0$.

Therefore, the temporal part is $g(t) = \sin(\omega t) = \sin(2\pi \nu t)$.

Combining the spatial and temporal parts, the equation of the standing wave is $y = A \sin\left(\dfrac{\pi x}{L}\right) \sin($2\pi \nu t)$.$