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I am forced to post this question here as physics SE refuses homework questions even if tagged with their (severely misleading) 'homework' tag. See my question at: https://physics.stackexchange.com/questions/105606/oscillating-motion-homework for more information.

I've got a 2,5kg object swinging without friction, back and forth on a spring and the time of revolution or T is 2 seconds.

When t=0, the item is in ''the position closest to the right side of equilibrium (x=+A)'' (trying to translate this, but it's weird even in my language).

When is the object at 3,5cm to the left of the equilibrium for the first time?

I've solved every part of this problem but this one and I'm stuck. I could use some hints to get me started.

What I understand is that the object is pretty much in equilibrium at t=0 and then it is dropped and I want to know when it is 3,5cm to the left but I have no idea how to perform the calculations.

Community
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Paze
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1 Answers1

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It doesn't sound like the object is in equlibrium at $t=0$, it sounds like the object is as far out of equilibrium as it can be (that is, as far away from the point where force is zero). Looks to me like you want to use the equation for the trajectory of a harmonic oscillator, $$ x(t) = A\cos(\omega t + \phi)$$ where $$\omega = \sqrt{\frac{k}{m}}=\frac{2\pi}{T}.$$ You know $m$ and $T$, so you should be able to back out $k$. Since the object is at $x=A$ at $t=0$, the phase shift $\phi$ is zero. Now find the smallest positive $t$ such that $x = 3.5$ cm. Since they didn't tell you the initial displacement, it should be in terms of $A$.

Neal
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  • Hm, isn't $A$ the amplitude? In other words the HIGHEST point the object can go? I was under the impression that equilibrium was in between the highest and lowest points (where the kinetic energy is maximized).

    Also, what does it mean for an object to be ''at the right apex of its trajectory''? Thank you for your answer, I'm glad to see that I am using the correct tools.

     – Paze Mar 29 '14 at 12:58
  • @Paze I was thinking in terms of a pendulum. Edited to be clearer. – Neal Mar 29 '14 at 13:16
  • Great. Thank you very much. – Paze Mar 29 '14 at 13:58