Simple Harmonic Motion — Question 9

Question 9

Acceleration changes fastest at the equilibrium crossing

For SHM, define jerk as $j=da/dt$. Let amplitude be $A=0.020\,\mathrm{m}$ and angular frequency $\omega=10\,\mathrm{rad/s}$.

1. Derive jerk in terms of velocity and find the maximum jerk magnitude.

2. At a positive-direction equilibrium crossing, find $v,a,j$. Contrast this with their values at the positive turning point.

Question 9 - Solution

Acceleration changes fastest at the equilibrium crossing

(a) Derivation and calculation

Differentiate the SHM acceleration law:$$a=-\omega^2x\quad\Rightarrow\quad j=\frac{da}{dt}=-\omega^2v.$$$$v_{\max}=A\omega,\qquad |j|_{\max}=\omega^2v_{\max}=A\omega^3.$$$$\boxed{|j|_{\max}=0.020(10)^3=20.0\,\mathrm{m/s^3}}.$$

(b) Result and interpretation

At the positive-direction equilibrium crossing,$$x=0:\quad v=+A\omega=0.20\,\mathrm{m/s},\quad a=0,\quad j=-20\,\mathrm{m/s^3}.$$At the positive turning point,$$x=A:\quad v=0,\quad a=-A\omega^2=-2.0\,\mathrm{m/s^2},\quad j=0.$$Thus maximum acceleration and maximum rate of acceleration occur at different phases.

Physical check

Jerk has units of acceleration divided by time; its sign is opposite the velocity for positive $\omega^2$.