Answer to Question #300174 in Physics for KeKo

Question

a particle moves with shm of period 4.0s and amplitude 4.0m. its displacement from the equilibrium position is x. find the time taken for it to travel: (a) from x=4.0m to x= 3.0m, (b) from x=-4.0m to x=3.0m, (c) from x=0 to x=3.0m, (d) from x=1.0 to x= 3.0m.

Answer 1

Given:

$T=4.0\:\rm s$

$A=4.0\:\rm m$

The equation of motion

$$x(t)=A\sin\left(\frac{2\pi}{T} t\right)=4.0\sin(0.5\pi t)$$ $$t=\frac{2}{\pi}\sin^{-1}\frac{x}{4.0}$$

(a)

$$\Delta t=\frac{2}{\pi}(\sin^{-1}\frac{4.0}{4.0}-\sin^{-1}\frac{3.0}{4.0})=0.46\:\rm s$$

(b)

$$\Delta t=\frac{2}{\pi}(\sin^{-1}\frac{3.0}{4.0}-\sin^{-1}\frac{-3.0}{4.0})=1.1\:\rm s$$

(c)

$$\Delta t=\frac{2}{\pi}(\sin^{-1}\frac{3.0}{4.0}-\sin^{-1}\frac{0}{4.0})=0.54\:\rm s$$

(d)

$$\Delta t=\frac{2}{\pi}(\sin^{-1}\frac{3.0}{4.0}-\sin^{-1}\frac{1.0}{4.0})=0.32\:\rm s$$