91Ó°ÊÓ

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)$\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\left(\mathrm{Ó\neg }t+\mathrm{Ï•}\right)$

Here, A is the amplitude of the wave,$\mathrm{Ó\neg }$ is the angular frequency, t is the time and$\mathrm{Ï•}$ 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,

$$\begin{gathered} \mathrm{Ó\neg }=2\mathrm{Î }f \\ \mathrm{Ó\neg }=2\mathrm{Î }\left(440Hz\right) \\ \mathrm{Ó\neg }=880\mathrm{Î }rad/s \end{gathered}$$

There is no phase angle, hence

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

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

$\begin{array}{r}{\mathrm{v}}_{max}=\mathrm{A}\mathrm{Ó\neg }\\ {\mathrm{v}}_{max}=\left(3.00×{10}^{−3}\mathrm{m}\right)(880\mathrm{Î °ù²¹»å}/\mathrm{s})\\ {\mathrm{v}}_{max}=8.294\mathrm{m}/\mathrm{s}\end{array}$

The maximum acceleration of the wave can be calculated as,

$\begin{array}{r}{\mathrm{a}}_{max}=\mathrm{A}{\mathrm{Ó\neg }}^2\\ {\mathrm{a}}_{max}=\left(3.00×{10}^{−3}\mathrm{m}\right)(880\mathrm{Î °ù²¹»å}/\mathrm{s}{)}^2\\ {\mathrm{a}}_{max}=22929\mathrm{m}/{\mathrm{s}}^2\end{array}$

$x\left(t\right)=\left(3.00×{10}^{−3}\mathrm{m}\right)\cos \left(\right(880\mathrm{Î °ù²¹»å}/\mathrm{s}\left)\mathrm{t}\right)$m

The third derivative equals the caution of the jerk,

$\begin{array}{r}j\left(t\right)=\frac{d^{3}x\left(t\right)}{d{t}^3}=\frac{d^3}{d{t}^3}\left(\left(3.00×{10}^{−3}\mathrm{m}\right)\cos \left(\right(880\mathrm{Î °ù²¹»å}/\mathrm{s}\left)\mathrm{t}\right)\right)\\ \mathrm{j}\left(\mathrm{t}\right)=\left(6.34×{10}^7\mathrm{m}/{\mathrm{s}}^3\right)\sin \left(\right(880\mathrm{Î °ù²¹»å}/\mathrm{s}\left)\mathrm{t}\right)\end{array}$

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

$\sin \left(\left(880\mathrm{Î }rad/s\right)t\right)$=1

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

Hence,

1. The equation of wave is$x\left(t\right)=\left(3.00×{10}^{−3}\mathrm{m}\right)\cos \left(\right(880\mathrm{Î °ù²¹»å}/\mathrm{s}\left)\mathrm{t}\right)$
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$j\left(t\right)=\left(6.34×{10}^7\mathrm{m}/{\mathrm{s}}^3\right)\sin \left(\right(880\mathrm{Î °ù²¹»å}/\mathrm{s}\left)\mathrm{t}\right)$ and the maximum jerk value is role="math" localid="1668093160860" $=6.34x{10}^7\mathrm{m}l{s}^3$.