2 The string shown in Active Figure 13.8 is driven at a frequency of
3 $5.00\U{Hz}$. The amplitude of the motion is $12.0\U{cm}$ and the
4 wave speed is $20.0\U{m/s}$. Furthermore, the wave is such that $y=0$
5 at $x=0$ and $t=0$. Determine \Part{a} the angular frequency
6 and \Part{b} wave number for this wave. \Part{c} Write an expression
7 for the wave function. Calculate \Part{d} the maximum transverse
8 speed and \Part{e} the maximum transverse acceleration of an element
14 This is just a unit conversion.
16 \frac{rad}{s}&=\frac{2\pi\U{rad}}{\text{cycle}}\cdot\frac{\text{cycles}}{s} \\
17 \omega &= 2\pi f = \ans{31.4\U{rad/s}}
21 Another units conversion
23 \frac{rad}{m} &= \frac{rad}{s} \cdot \frac{s}{m} \\
24 k &= \frac{\omega}{v} = \ans{1.57\U{rad/m}}
28 Now we get to plug in those values for $k$ and $\omega$.
30 y(x,t) = A \sin(kx-\omega t)
31 = 12.0\U{cm}\cdot\sin(1.57\U{rad/m}\cdot x - 31.4\U{rad/s}\cdot t)
35 Differentiating with respect to time pulls out a chain-rule $\omega$.
37 \p({\pderiv{t}{y}})_\text{max} = A\omega = 12.0\U{cm}\cdot 31.4\U{1/s}
42 Differentiating again with respect to time pulls out another $\omega$.
44 \p({\npderiv{2}{t}{y}})_\text{max} = A\omega^2
45 = 12.0\U{cm}\cdot (31.4\U{1/s})^2 = \ans{118\U{m/s$^2$}}