Spring Constant given mass, and equivalent mass in free fall from fixed height.

AI Thread Summary
The discussion revolves around calculating the spring constant for a system involving two 54kg blocks, one resting on a spring and the other in free fall. The blocks are released simultaneously, with the block on the left hitting the table while the block on the right reaches an instantaneous rest. To find the spring constant, the time of free fall is calculated, equating it to the time it takes for the spring-mounted block to reach its first instantaneous rest, which occurs at a quarter of the oscillation period. The initial calculations led to a spring constant of 2.176 N/m, but this was deemed incorrect due to a misunderstanding of the point of balance in the oscillation. Clarification was sought regarding the definitions of "instantaneous rest" and the forces acting on the spring-mounted block.
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Homework Statement


Two 54kg blocks are held 30cm above a table. As shown in the figure, one of them is just touching a 30cm long spring. The blocks are released at the same time. The block on the left hits the table at exactly the same instant as the block on the right first comes to an instantaneous rest.

What is the spring constant?

Homework Equations


Delta L = mg/k


The Attempt at a Solution


I think the solution might start with finding the time it takes for the free fall block to hit the surface, which is the same as the block mounted on the spring. Then we can use to some how find the compression length. But that's as far as I have got.
 
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Yes, right. First of all, you can calculate the time that the block without a spring in its belly needs in order to reach the floor and equate this with the time that the other block needs in order to first come to instantaneous rest. Of course the second block's motion is going to be a part of a simple harmonic oscillation, so the time until first instantaneous rest is a quarter of the oscillation's period, meaning t = T/4. Use that equation to calculate the spring constant.
 
Awesome, so this is what I did from your suggestion:

d=vi * t + 0.5 * a * t^2
0.3m = 0 + 0.5 * 9.8 * t^2
t = 0.2473 seconds

0.2573 * 4 = 0.9897 = period

frequency = 1 / period
frequency = 1.01 hz

frequency = (1/2pie) * sqrt(k/mass)
1.01 = (1/2pie) * sqrt(k/0.054kg)
k = 2.176 N/m

But however, my professor told me this was wrong... please let me know where I made mistakes.
 
Oh yea, that's true, my bad. You see, the point of balance for the oscillation of the second block isn't the point that it initially is, where it touches the spring, but a little below that, so that the force from the spring is opposite to the block's weight. The second block first reaches that point and T/4 later it reaches it's first instantaneous rest point.There's a chance we're talking about different things here. When saying instantaneous rest point you mean i) the point where ΣF=0 for the second block or ii) the point where v=0 (velocity) for the second block?? Because I'm not perfectly familiar with the terms in english.
 
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