# More Simple Harmonic Oscillations

1. Nov 8, 2006

### ubbaken

An object in simple harmonic motion oscillates with a period of 4.00 s and an amplitude of 9.08 cm. How long does the object take to move from x=0.00 cm to x=5.07 cm?

I set up my eqn like this: 0.0908cos(ωt)=0.0507
cos(ωt)=0.583
ωt=56.1
then with ω=90deg I get 0.623s which is slightly higher than I think the answer would be, and is ultimately incorrect.

I've tried working the problem out a number of times, in rad and degs to make sure I'm not making an error there, and I get the same answer.

The velocity of an object in simple harmonic motion is given by v(t)= -(0.250 m/s)sin(17.0t + 1.00π), where t is in seconds. What is the first time after t=0.00 s at which the velocity is -0.120 m/s?

for this one it seems like I just have to work it through to find t. When I do so, I get down to 17.0t+1.00π=28.7 (from using sin^-1=-.120/-.250) and then there is no answer that fits (0<t<5 approx.). Plus, in trying the answers, they are wrong.

I'm pretty spent on these, any and all help would be appreciated.

Last edited by a moderator: Apr 22, 2017
2. Nov 8, 2006

### Mindscrape

You have two condtions x(0)=0, and x(t)=5.07cm. To use the equation that $$x(t)=Acos(\omega t + \phi)$$ you have to first find the phase constant. So you need to make a statement like
$$x(t)=Acos(\omega t + \phi)=0$$

It is pretty obvious for this one because you essentially want to just start with a sine function. But you can solve and find the phase. Then you want to use that phase you found to solve when $$x(t)=Acos(\omega t + \phi)=5.07cm$$. Another problem I see with what you did is you are saying the angular frequency has units of degrees? Remember that $$\omega = \frac{2 \pi}{T}$$ where T is the period, and has units of s^-1.

The second one is very similar, if you can get the first one then you can get the second one.