The Force Law for Simple Harmonic Motion — Question 4

Question 4

An accelerating support changes the center seen in its own frame

A horizontal spring of constant $k$ is fixed inside a cart accelerating rightward at constant $a_c$. A mass $m$ slides without friction relative to the cart. Let $x$ be its rightward displacement from the spring's unstretched position in the cart frame. Treat the spring as bilateral.

1. Derive the cart-frame equation, equilibrium and small-oscillation frequency.

2. If the cart begins this acceleration abruptly while the mass is at rest relative to it at $x=0$, find the ensuing relative amplitude. What quantity remains continuous at the switch?

Question 4 - Solution

An accelerating support changes the center seen in its own frame

(a) Derivation and calculation

In the accelerating cart frame,$$m\ddot x=-kx-ma_c,\qquad0=-kx_e-ma_c\Rightarrow\boxed{x_e=-\frac{ma_c}{k}}.$$$$q=x-x_e:\quad m\ddot q=-kq\Rightarrow\boxed{\omega=\sqrt{\frac{k}{m}}}.$$

(b) Result and interpretation

No impulsive force acts when acceleration changes:$$x(0^+)=x(0^-)=0,\qquad\dot x(0^+)=\dot x(0^-)=0.$$$$q(0)=\frac{ma_c}{k},\quad\dot q(0)=0\Rightarrow\boxed{A=\frac{ma_c}{k}}.$$$$x(t)=-\frac{ma_c}{k}\left[1-\cos\!\left(\sqrt{k/m}\,t\right)\right].$$The center shifts left; the frequency depends on stiffness and mass.

Physical check

At switch-on, $\ddot x(0)=-a_c$, consistent with a mass initially tending to retain its laboratory velocity.