Spring with mass on horizontal surface

  • Thread starter teme92
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  • #1
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Homework Statement



A 2.00 kg object on a horizontal frictionless surface is attached to a spring with
a spring constant 1000 N/m. The object is displaced from equilibrium 50.0 cm
horizontally and given an initial velocity of 4.0 m/s away from the equilibrium
position.

(a) What is the frequency of motion?

(b) What is the initial potential energy of the block-spring system?

(c) What is the initial kinetic energy?

(d) What is the amplitude of oscillation?


Homework Equations



F = -kx
F = ma
[itex]\omega[/itex] = √(k/m)
T = 1/f
f = [itex]\omega[/itex]/2[itex]\pi[/itex]
PE = 0.5kx2
KE = 0.5mv2

The Attempt at a Solution



(a) So I said:

T = 1/f → T = 2[itex]\pi[/itex]/f → T = 2[itex]\pi[/itex]√(m/k)

I subbed in my values and got T = 0.28s

Then I got f from f = 1/T and got f = 3.57s-1

(b) From PE = 0.5kx2 I got PE = 125J

(c) From KE = 0.5mv2 I got KE = 16J

(d) Its the amplitude part here that I got stuck. Could anyone clarify if my methods for the other parts were correct and point me some direction for part (d). Any help is much appreciated.
 
Last edited:

Answers and Replies

  • #2
BvU
Science Advisor
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(c) is weird. check.
for (d) you can use the fact that at displacement from equilibrium = amplitude value, the movement changes direction, so kinetic energy = 0.
 
  • #3
185
2
Hey BvU thanks for the help. (c) should be 16 I forgot to square the 4. Ok so I can say 0.5mv2 = 0 but how to I get an 'A' value from this?
 
  • #4
185
2
Wait I think I have it now.

So PE + KE = 0.5kA2

KE = 0
PE = 125

So work to get A from there?

I got A = 0.5m which is the same as x so I've a feeling this is wrong.
 
Last edited:
  • #5
185
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So I have another attempt :tongue:

The work done decreases the kinetic energy to zero so can I say:

0.5kA2 = 0.5mv2

Solving for A = 0.179m
 
  • #6
107
13
The spring was given an initial velocity + an initial displacement. So, you have factor both of these things if you are using the ##1/2kA^2## equation, remember that this is equal to the total mechanical energy in your system. It would make no sense for your spring to have a smaller amplitude if it would have more energy in the system as opposed to having no initial velocity.
 
  • #7
185
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Ok so how do I factor in the velocity and displacement? I thought I was with the 'v' in 0.5mv2 and the x in 0.5kx2.
 
  • #8
107
13
Ok so how do I factor in the velocity and displacement? I thought I was with the 'v' in 0.5mv2 and the x in 0.5kx2.
What is the initial energy put into your system? Don't even worry about the amplitude yet, just figure out the total energy in the system based on the initial displacement from equilibrium and the initial velocity.
 
  • #9
185
2
E = KE + PE = 125 + 16 = 141J

Do I let that equal to the work energy equation then?
 
  • #10
107
13
If that's your total mechanical energy, and your know that ##1/2kA^2 = E_{total}## , then you have all you need it looks like.
 
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  • #11
185
2
Yep I have the answer worked out, thanks a million for the help. Studying for my finals and this really speeds up the process when I'm stuck.
 
  • #12
107
13
Glad to hear it, good luck on your finals!
 

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