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

In summary, the two blocks, one without a spring and one with a spring, are released at the same time. The block with a spring hits the table at the same instant as the block without a spring. The spring constant is 0.9897.
  • #1
wzx7410
2
0

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.
 
Physics news on Phys.org
  • #2
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 [itex]t = T/4[/itex]. Use that equation to calculate the spring constant.
 
  • #3
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.
 
  • #4
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.
 
Last edited:

What is the spring constant?

The spring constant is a measure of the stiffness or rigidity of a spring. It is defined as the force required to stretch or compress a spring by a certain distance.

How is spring constant related to mass?

The spring constant is not directly related to mass. However, the force exerted by a spring on an object is directly proportional to the mass of the object. This means that as the mass increases, the force exerted by the spring also increases.

What is the equivalent mass in free fall from a fixed height?

The equivalent mass in free fall from a fixed height is the mass that would fall with the same acceleration as an object attached to a spring. It is calculated by dividing the force exerted by the spring by the acceleration due to gravity.

How does the spring constant affect the free fall equivalent mass?

The spring constant does not affect the free fall equivalent mass. This is because the mass of the object attached to the spring remains constant. The only factors that affect the equivalent mass in free fall are the force exerted by the spring and the acceleration due to gravity.

Can the spring constant be changed?

Yes, the spring constant can be changed by altering the physical properties of the spring, such as its material, length, and thickness. The spring constant can also be changed by adjusting the number of springs used in a system.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
290
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
30
Views
719
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
919
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
2K
Back
Top