By sinusoidally vibrating one end of a long horizontal string up...

Given the amplitude $A = 1.0 \text{ cm} = 0.01 \text{ m}$.

Since the displacement reaches zero $200$ times per second and a sinusoidal wave crosses zero twice per cycle, the frequency is $f = \dfrac{200}{2} = 100 \text{ Hz}$.

The angular frequency is $\omega = 2\pi f = 200\pi \text{ rad s}^{-1}$.

The wave speed is given by $v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{90}{0.10}} = 30 \text{ m s}^{-1}$.

The wave number is $k = \dfrac{\omega}{v} = \dfrac{200\pi}{30} = \dfrac{20\pi}{3} \text{ m}^{-1}$.

The wave propagates in the positive $x$-direction. At $t = 0$ and $x = 0$, the particle is at its positive extreme, so the wave equation is $y(x,t) = A \cos(\omega t - kx)$.

We need to find the velocity and acceleration at $x = 50 \text{ cm} = 0.5 \text{ m}$ and $t = 10 \text{ ms} = 0.01 \text{ s}$.

The phase angle is $\theta = \omega t - kx = (200\pi)(0.01) - \left(\dfrac{20\pi}{3}\right)(0.5) = 2\pi - \dfrac{10\pi}{3} = -\dfrac{4\pi}{3}$.

The particle velocity is $v_p = \dfrac{\partial y}{\partial t} = -A\omega \sin(\omega t - kx)$.

Substituting the values, $v_p = -(0.01)(200\pi) \sin\left(-\dfrac{4\pi}{3}\right) = -2\pi \left(\dfrac{\sqrt{3}}{2}\right) = -\sqrt{3}\pi \text{ m s}^{-1}$.

The particle acceleration is $a_p = \dfrac{\partial^2 y}{\partial t^2} = -A\omega^2 \cos(\omega t - kx)$.

Substituting the values, $a_p = -(0.01)(200\pi)^2 \cos\left(-\dfrac{4\pi}{3}\right) = -400\pi^2 \left(-\dfrac{1}{2}\right) = 200\pi^2 \text{ m s}^{-2}$.