Answer to Question #101709 in Physics for James

Question

An object in SHM oscillates with a period of 4.0 s and an amplitude of 14 cm. How long does the object take to move from x = 0.0 cm to x = 5.9 cm.
(The answer should be 0.28 seconds, but I keep getting 1.7s )
w=2pi/T = pi/2 and x(t)=Acos(wt+phi) so x(t)=14cos(pi*t/2 + phi) .... phi=+ or - pi/2
Am I supposed to use the positive or negative value? If is use the negative value, substitute x(t) for 5.9 and solve for t, I get 1.7seconds, but if I use the positive value, substitute x(t) for 5.9 and solve for t, I get Negative 0.28seconds which doesn't make sense. What am I doing wong?

Answer 1

The object’s position as a function of time is

$$x=A\cos(\omega t+\phi)$$

$$x(0)=0\to \phi=\pm \frac{\pi}{2}$$

Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between $-\pi$ and 0 radians. Thus,

$$\phi=- \frac{\pi}{2}$$

$$\omega=\frac{2\pi}{T}=\frac{2\pi}{4}=\frac{\pi}{2}\frac{rad}{s}$$

$$x=A\cos(\frac{\pi}{2}t-\frac{\pi}{2})=A\sin{\frac{\pi}{2}t}$$

So,

$$0.059=0.14\sin{\frac{\pi}{2}t}$$

$$t=0.28\ s$$