Spring motion

  • Thread starter Psyguy22
  • Start date
  • #1
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Homework Statement


When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled down from this position and released. It undergoes 57 oscillations in 39.0 s. What was the distance d?


Homework Equations


F=-Kx
F=mg
T=2pi*sqrt(K/m)


The Attempt at a Solution


I started by putting the first two equations equal.
[itex] -K x= m g [/itex]
Solving for x you get
[itex] [1] (m g) /-K= x [/itex]
Then I solved for m/K in the third equation.
[itex][2] (2 π / T )^2=m / K [/itex]
Then I solved for osc/s which came to be 1.46 osc/s
Substituting {2} for {1} I came up with
[itex] g*(- 2 π/ T)^2= x [/itex] BUt i feel like this equation is wrong. Any help?
 

Answers and Replies

  • #2
Do not set the first two equations equal to each other, they are not equal. The two forces acting on the mass, the weight and spring force, add to give the total force which will be zero when x=d.
 
  • #3
Doc Al
Mentor
45,140
1,439
T=2pi*sqrt(K/m)
This is incorrect. You have K and m reversed.

I started by putting the first two equations equal.
[itex] -K x= m g [/itex]
Solving for x you get
[itex] [1] (m g) /-K= x [/itex]
Here the problem is that minus sign. Set the magnitudes of the forces equal.

Fix those problems and you'll be OK.
 
  • #4
62
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Ok i fixed the two mistakes, but I'm still coming up with the wrong answer. I think I have something wrong with the peiod but am unsure how to fix it
 
  • #5
Doc Al
Mentor
45,140
1,439
Ok i fixed the two mistakes, but I'm still coming up with the wrong answer. I think I have something wrong with the peiod but am unsure how to fix it
The period is the time it takes for one oscillation. Above you solved for the osc/sec, which is the frequency, not the period.
 

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