A wave is described by the equation y = (1.0 mm) π ( x 2.0 cm - t...

The given wave equation is y = (1.0 \text{ mm}) \sin \left(\dfrac{\pi x}{2.0} -100 \pi t\right)$.<br><br>The particle velocity$v$is given by the partial derivative of$y$with respect to time$t$:<br><br>$v = $\dfrac{\partial y}{\partial t} = ($1.0 \text{ mm}) \cos \left(\dfrac{\pi x}{2.0} -100 \pi t\right) \times (-100 \pi \text{ s}^{-1})

v = -100 \pi \cos \left(\dfrac{\pi x}{2.0} -100 \pi t\right) \text{ mm s}^{-1}$<br><br>At$x = 1.0 \text{ cm}$, the argument of the cosine becomes \dfrac{\pi}{2} -$100 \pi t$. Using$\cos\left(\dfrac{\pi}{2} - \theta\right) = \sin \theta, the velocity simplifies to:

$v = -100 \pi \sin(100 \pi t) \text{ mm s}^{-1} = -10 \pi \sin(100 \pi t) \text{ cm s}^{-1}$

The speed is the magnitude of the velocity, $|v| = 10 \pi |\sin(100 \pi t)| \text{ cm s}^{-1}$.

At $t = 0.011 \text{ s}$:

$|v| = 10 \pi |\sin(1.1 \pi)| = 10 \pi \sin(18^\circ) = 10 \pi \left(\dfrac{\sqrt{5}-1}{4}\right) \approx$9.7 \text{ cm s}^{-1}$<br><br>At$t = 0.012 \text{ s}$:<br><br>$|v| = 10 \pi |\sin(1.2 \pi)| = 10 \pi \sin(36^\circ) = 10 \pi \left(\dfrac{\sqrt{10-2\sqrt{5}}}{4}\right) \approx$18.5 \text{ cm s}^{-1}$

At $t = 0.013 \text{ s}$:

$|v| = 10 \pi |\sin(1.3 \pi)| = 10 \pi \sin(54^\circ) = 10 \pi \left(\dfrac{\sqrt{5}+1}{4}\right) \approx$25.4 \text{ cm s}^{-1}$<br><br>The speeds are approximately$9.7 \text{ cm s}^{-1}$,$18 \text{ cm s}^{-1}$, and$25 \text{ cm s}^{-1}$.<br><br>Answer:$9.7 \text{ cm s}^{-1}$,$18 \text{ cm s}^{-1}$,$25 \text{ cm s}^{-1}