A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 mm and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

Simple harmonic vibratory motion is a type of mechanical displacement described in Newtonian physics as the relative change in position of a particle moving in a periodic path. This implies that the particle will pass through its resting point in equal periods as a swing.

a)$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(\omega t+\varphi \right)$

Here, A is the amplitude of the wave,$\omega$ is the angular frequency, t is the time and$\varphi$ is the phase angle.

According to the information given,

Frequency f=440Hz and Amplitude A=3.00mm=3.00×10−3m.

The angular frequency can be easily found through the frequency value,

$$\begin{gathered} \omega =2\Pi f \\ \omega =2\Pi \left(440Hz\right) \\ \omega =880\Pi rad/s \end{gathered}$$

There is no phase angle, hence

$x\left(t\right)=\left(3.00\times {10}^{-3}\right)\cos \left(\left(880\Pi rad/s\right)t\right)$

b) The maximum speed of the wave can be calculated as,

$\begin{array}{r}{\mathrm{v}}_{max}=\mathrm{A}\omega \\ {\mathrm{v}}_{max}=\left(3.00\times {10}^{-3}\mathrm{m}\right)(880\mathrm{\Pi rad}/\mathrm{s})\\ {\mathrm{v}}_{max}=8.294\mathrm{m}/\mathrm{s}\end{array}$

The maximum acceleration of the wave can be calculated as,

$\begin{array}{r}{\mathrm{a}}_{max}=\mathrm{A}{\omega }^2\\ {\mathrm{a}}_{max}=\left(3.00\times {10}^{-3}\mathrm{m}\right)(880\mathrm{\Pi rad}/\mathrm{s}{)}^2\\ {\mathrm{a}}_{max}=22929\mathrm{m}/{\mathrm{s}}^2\end{array}$

$x\left(t\right)=\left(3.00\times {10}^{-3}\mathrm{m}\right)\cos \left(\right(880\mathrm{\Pi rad}/\mathrm{s}\left)\mathrm{t}\right)$m

The third derivative equals the caution of the jerk,

$\begin{array}{r}j\left(t\right)=\frac{d^{3}x\left(t\right)}{d{t}^3}=\frac{d^3}{d{t}^3}\left(\left(3.00\times {10}^{-3}\mathrm{m}\right)\cos \left(\right(880\mathrm{\Pi rad}/\mathrm{s}\left)\mathrm{t}\right)\right)\\ \mathrm{j}\left(\mathrm{t}\right)=\left(6.34\times {10}^7\mathrm{m}/{\mathrm{s}}^3\right)\sin \left(\right(880\mathrm{\Pi rad}/\mathrm{s}\left)\mathrm{t}\right)\end{array}$

The maximum value of jerk will be at the time when,

$\sin \left(\left(880\Pi rad/s\right)t\right)$=1

The max value of$j\left(t\right)=6.34x10ml{s}^3$.

Hence,

1. The equation of wave is$x\left(t\right)=\left(3.00\times {10}^{-3}\mathrm{m}\right)\cos \left(\right(880\mathrm{\Pi rad}/\mathrm{s}\left)\mathrm{t}\right)$
2. The max speed and acceleration of wave at the center is${\mathrm{v}}_{max}=8.294m/s$ and$a_{max}=22929\mathrm{m}l{s}^2$ respectively.
3. The equation of jerk is$j\left(t\right)=\left(6.34\times {10}^7\mathrm{m}/{\mathrm{s}}^3\right)\sin \left(\right(880\mathrm{\Pi rad}/\mathrm{s}\left)\mathrm{t}\right)$ and the maximum jerk value is role="math" localid="1668093160860" $=6.34x{10}^7\mathrm{m}l{s}^3$.