Simple Harmonics - Kind of wierd

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The discussion revolves around solving a harmonic motion problem involving a mass on a spring, specifically finding the position of the mass at t = 1.00s using the provided displacement equation. Initially, the user attempted to substitute the time directly into the equation but received an incorrect result. It was clarified that the calculations should be performed in radians rather than degrees, which resolved the issue. The correct position was confirmed to be obtained when using radians. The conversation highlights the importance of using the appropriate angle measurement in harmonic motion calculations.
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[SOLVED] Simple Harmonics - Kind of wierd

Homework Statement



Question:
A 1.30 kg mass on a spring has displacement as a function of time given by the equation:

x\left( t \right) = \left( {7.40\;{\rm{ cm}}} \right)\,\cos\:\left[ {\left( {4.16\;{\rm{ s}}^{ - 1} } \right)t - 2.42} \right].

Find the position of the mass at t = 1.00s.

Homework Equations



?

The Attempt at a Solution



Ok so since this is a position equation, i figured that i would just plugin 1 for t and get the answer. But that seems to be wrong and i can't figure out why. (i did change 7.4 cm to m).

I figured out other parts for this equation to be:
T = 1.51s
k = 22.5 N/m
Vmax = .308 m/s
Fmax = 1.67 N

Any help would be great!

Thanks!
 
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Hello,

Well, if you have a position equation that is only a function of time, then yes, you should be able to plug in the time and get the position.

So

Is this an equation that was given to you, or is this an equation you derived from the problem? You may have made a mistake in an earlier step. What exactly did the original problem say?
 
Its an equation given to me.

Problem said exactly what i posted. Everything was given to me and everything is correct.

When i solve this: i get .073965879

Can anyone confirm this? I tried that and it was wrong. (my calculator is in degrees mode - i think that's right)
 
Hello,

Well, I get the same number in degrees mode, but you probably want to be working in radians, as this is the unit typically used for these types of things. Try radians.

What answer are you looking to get?
 
O ok it works in radians.

Thanks!
 
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