### Challenge Problems

A copper wire has a radius of $200\phantom{\rule{0.2em}{0ex}}\text{\mu m}$ and a length of 5.0 m. The wire is placed under a tension of 3000 N and the wire stretches by a small amount. The wire is plucked and a pulse travels down the wire. What is the propagation speed of the pulse? (Assume the temperature does not change: $\left(\rho =8.96\frac{\text{g}}{{\text{cm}}^{3}},Y=1.1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{11}\frac{\text{N}}{\text{m}}\right).)$

A pulse moving along the *x* axis can be modeled as the wave function $y\left(x,t\right)=4.00\phantom{\rule{0.2em}{0ex}}\text{m}{e}^{\text{\u2212}{\left(\frac{x+(2.00\phantom{\rule{0.2em}{0ex}}\text{m/s})t}{1.00\phantom{\rule{0.2em}{0ex}}\text{m}}\right)}^{2}}.$ (a)What are the direction and propagation speed of the pulse? (b) How far has the wave moved in 3.00 s? (c) Plot the pulse using a spreadsheet at time $t=0.00\phantom{\rule{0.2em}{0ex}}\text{s}$ and $t=3.00\phantom{\rule{0.2em}{0ex}}\text{s}$ to verify your answer in part (b).

A string with a linear mass density of $\mu =0.0085\phantom{\rule{0.2em}{0ex}}\text{kg/m}$ is fixed at both ends. A 5.0-kg mass is hung from the string, as shown below. If a pulse is sent along section *A*, what is the wave speed in section *A* and the wave speed in section *B*?

Consider two wave functions ${y}_{1}\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t\right)$ and ${y}_{2}\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx+\omega t+\varphi \right)$. What is the wave function resulting from the interference of the two wave? (*Hint:* $\text{sin}\left(\alpha \pm \beta \right)=\text{sin}\phantom{\rule{0.2em}{0ex}}\alpha \phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}\beta \pm \text{cos}\phantom{\rule{0.2em}{0ex}}\alpha \phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\beta $ and $\varphi =\frac{\varphi}{2}+\frac{\varphi}{2}$.)

The wave function that models a standing wave is given as ${y}_{R}\left(x,t\right)=6.00\phantom{\rule{0.2em}{0ex}}\text{cm}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(3.00\phantom{\rule{0.2em}{0ex}}{\text{m}}^{\mathrm{-1}}x+1.20\phantom{\rule{0.2em}{0ex}}\text{rad}\right)$ $\text{cos}\left(6.00\phantom{\rule{0.2em}{0ex}}{\text{s}}^{\mathrm{-1}}t+1.20\phantom{\rule{0.2em}{0ex}}\text{rad}\right)$. What are two wave functions that interfere to form this wave function? Plot the two wave functions and the sum of the sum of the two wave functions at $t=1.00\phantom{\rule{0.2em}{0ex}}\text{s}$ to verify your answer.

Consider two wave functions ${y}_{1}\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-\omega t\right)$ and ${y}_{2}\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx+\omega t+\varphi \right)$. The resultant wave form when you add the two functions is ${y}_{R}=2A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx+\frac{\varphi}{2}\right)\text{cos}\left(\omega t+\frac{\varphi}{2}\right).$ Consider the case where $A=0.03\phantom{\rule{0.2em}{0ex}}{\text{m}}^{\mathrm{-1}},$ $k=1.26\phantom{\rule{0.2em}{0ex}}{\text{m}}^{\mathrm{-1}},$ $\omega =\pi \phantom{\rule{0.2em}{0ex}}{\text{s}}^{\mathrm{-1}}$, and $\varphi =\frac{\pi}{10}$. (a) Where are the first three nodes of the standing wave function starting at zero and moving in the positive *x* direction? (b) Using a spreadsheet, plot the two wave functions and the resulting function at time $t=1.00\phantom{\rule{0.2em}{0ex}}\text{s}$ to verify your answer.