How to Calculate Time for a Mass on a Spring to Reach Equilibrium Again?

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To calculate the time for a mass on a spring to reach equilibrium after being stretched, the spring constant (k) is first determined using the formula F = kx. The correct approach involves calculating the angular frequency (omega) with omega = sqrt(k/m) and then finding the frequency (f) and period (T) using f = omega/2pi and T = 1/f. It's crucial to note that the time to reach the new equilibrium position is only one-fourth of the period, as the mass moves through a quarter of its oscillation cycle. The initial calculations regarding force and acceleration are not applicable since the motion involves oscillation rather than constant acceleration. Understanding these concepts is essential for solving the problem accurately.
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Q:
A mass of 1.64 kg stretches a vertical spring 0.300 m. If the spring is stretched an additional 0.123 m and released, how long does it take to reach the (new) equilibrium position again?

What I tried:
Found k using F=kx (mg=kx)
using that k, I calculated the force in the extended spring (F=k(0.123))
using that force, I calculated the acceleration of the mass (F=ma)
finally, using x=(1/2)at^2 I calculated t.

I must be missing something obvious b/c this doesn't seem like a difficult problem... :confused:
 
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Nope,it's incorrect.The motion is not with constant acceleration.U need to find the period of vibration.

Daniel.
 
Yeah that's what I thought...

So I calculated k in the first step above.
Then I used omega = sqrt(k/m) to get omega.
Then I used f = omega/2pi
Then period = 1/f and it's still wrong. Any other mistakes I made?

Thanks for the help :smile:
 
Yes,the time required is only 1/4-th of the period...Can u see why?

Daniel.
 
Wow hey thanks a lot. I guess brain just isn't working today. :smile:
 

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