Simple Harmonic Motion, Finding Time

AI Thread Summary
A 0.400 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm. To find the time interval for the object to move from x = 0 to x = 4.00 cm, the correct approach involves using the sine function instead of the cosine function, as the cosine function assumes the object starts at the maximum amplitude. The solution requires calculating the angular frequency (ω) and ensuring consistent units during calculations. After correcting the approach, the time can be accurately determined. The discussion highlights the importance of selecting the appropriate trigonometric function based on the initial conditions of the motion.
Ithryndil
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[SOLVED] Simple Harmonic Motion, Finding Time

Homework Statement



A 0.400 kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.0 cm.

Calculate the time interval required for the object to move from x = 0 to x = 4.00 cm.


Homework Equations


The relevant equations for this should be:

x = Acos(ωt)

w = \sqrt{k/m}


The Attempt at a Solution



You should be able to solve the above equation for t...

cos(ωt) = x/A
ωt = arccos(x/A)
t = arccos(x/A)/\sqrt{k/m}

Plugging in the following values:

x = 4 cm
A = 10 cm
k = 8 N/m
m = 0.400 kg

I get that t = .259s which is not right. Where am I going wrong?
 
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Just as a quick notice, your "plug in" values have inconsistent units... you have both cm and m
Would that fix the problem?
 
That shouldn't matter because the units for the arccos(x/A) should simply cancel each other out.
 
Hi Ithryndil,

The problem is your trig function cos(w t). That function at t=0 indicates that the particle is at the positive amplitude, so you found the time to go from x=10 cm to x=4 cm.

If instead you use a sine function, since it is zero at t=0, you should be able to follow the rest of your procedure. (Or you could put a phase shift \phi into your cosine function to make it act like a sine function.)

But whenever your problem depends on the oscillator being at a specific position at t=0 (and perhaps a specific velocity direction), you must make sure that your trig function has that same behavior.
 
Ok. Thank you for your help. That worked.
 
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