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

The amplitude of the wave is given as $A = 1.0 \text{ cm}$.

Since the displacement of a particle in simple harmonic motion reaches zero twice in one complete cycle, the frequency $f$ of the wave is:

$f = \dfrac{200}{2} = 100 \text{ Hz}$

The time period $T$ is:

$T = \dfrac{1}{f} = \dfrac{1}{100} = 0.01 \text{ s}$

The wave speed $v$ is determined by the tension $F and the linear mass density$\mu$:<br><br>$v = \sqrt{\dfrac{F}{\mu}} = \sqrt{\dfrac{90}{0.10}} = \sqrt{900} = 30 \text{ m s}^{-1} = 3000 \text{ cm s}^{-1}$<br><br>The wavelength$\lambda$is:<br><br>$\lambda = \dfrac{v}{f} = \dfrac{3000}{100} = 30 \text{ cm}$<br><br>The general equation for a wave travelling in the positive$x$-direction is:<br><br>$y(x,t) = A \cos \left(2\pi\left(\dfrac{x}{\lambda} - \dfrac{t}{T}\right) + \phi\right)

At $x = 0$ and $t = 0$, the displacement is at its positive extreme, so $y(0,0) = A$. This gives $\cos(\phi) = 1$, which implies $\phi = 0$.

Substituting the values of $A$, $\lambda$, and $T$, we get:

y = (1.0 \text{ cm}) \c\cos 2\pi \left[\dfrac{x}{30 \text{ cm}} - \dfrac{t}{0.01 \text{ s}}\right]