Hooke's Law: Spring, Distance and Time

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Homework Help Overview

The problem involves a mass moving on a frictionless air track and interacting with a spring according to Hooke's Law. Participants are tasked with determining the maximum compression of the spring and the total time for the mass to return to its starting point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using energy conservation principles to find the maximum compression of the spring and the time taken for the mass to return. There are attempts to apply equations related to kinetic energy and spring potential energy.

Discussion Status

Some participants have provided guidance on correcting the approach to the energy conservation equation for part (a). Others are exploring different methods for calculating the time in part (b), noting the need to consider the motion before and after the spring compression.

Contextual Notes

Participants are working under the constraints of the problem statement, including the initial conditions of the mass and spring constants. There is an acknowledgment of the need to separate the motion into distinct phases for accurate analysis.

kgal
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Homework Statement


A mass M = 0.454 kg moves with an initial speed v = 2.88 m/s on a level frictionless air track. The mass is initially a distance D = 0.250 m away from a spring with k = 876 N/m, which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance d, before reversing direction. after bouncing off the spring, the mass travels with the same speed v, but in the opposite direction.
a. Determine the maximum distance that the spring is compressed (0.0656 m).
b. Find the total elapsed time until the mass returns to its starting point (0.245 s).


Homework Equations



F = -kx



The Attempt at a Solution



a. Tried to use energy conservation equation but couldn't get the correct answer.
1/2mv0^2 + 1/2kd^2 + 1/2m1v^2 = 0.
got d = √(2mv^4/k) which got me an incorrect answer.

b. have no idea where to begin
 
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kgal said:

The Attempt at a Solution



a. Tried to use energy conservation equation but couldn't get the correct answer.
1/2mv0^2 + 1/2kd^2 + 1/2m1v^2 = 0.
got d = √(2mv^4/k) which got me an incorrect answer.

The line in red is incorrect. The change in KE = work done by the spring. So you should have

1/2m(v2-v20 = 1/2 kd2

And you know that v=0. This should get you the correct value for d.
 
Thanks a lot!
 
for section b,
I was thinking of using the equation x(t) = Acos(sqrt(k/m)t) where x(t) = 0.25 m, k from the answer in section a and the given mass, but can't get anywhere when trying to solve for t...
 
kgal said:
for section b,
I was thinking of using the equation x(t) = Acos(sqrt(k/m)t) where x(t) = 0.25 m, k from the answer in section a and the given mass, but can't get anywhere when trying to solve for t...

The mass starts out 0.250m away from the spring... it's not even in contact with it. It first travels that distance at its initial constant velocity before coming in contact with the spring. Then it compresses the spring by the distance found in part (a) and is subsequently launched back along its initial trajectory.

You'll have to divide the problem into separate parts corresponding to when it is in contact with and not in contact with the spring. When it's in contact with the spring you can use properties of a mass-spring system that you should know.
 

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