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Force on a vertical spring

  • Thread starter clope023
  • Start date
  • #1
987
123
1. Homework Statement

A force of magnitude 39.5N stretches a vertical spring a distance 0.256m.

a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.10s?

got 4.73kg

b) If the amplitude of the motion is 5.00×10−2m and the period is that specified in part (a), where is the object at a time 0.320s after it has passed the equilibrium position, moving downward? (Take the upward direction positive.)

c) What force (magnitude) does the spring exert on the object when it is a distance 2.50×10−2m below the equilibrium position, moving upward?

d) direction of part c's force?

2. Homework Equations

T = 2pi/omega

omega = 2pi/T

x = Acos(omegat+phi)

F = kx

F/x = k = 154.3N

T = 2pi(m/k)^1/2


3. The Attempt at a Solution

b) x = Acos((2pi/1.10)*.320) = .04998 ~ .05m - incorrect

c) 1. F = kx = (154.3)(2.5x10^-2) = 3.9 - incorrect

2. F = k[tex]\int[/tex]xdx (from 5x10^-2 to 2.5x10^-2)

= -.1446m - incorrect

I don't know why the equations aren't working for me, any help is appreciated.
 

Answers and Replies

  • #2
987
123
I believe it has something to do with the period after the equilibrium position, but am not sure.
 
  • #3
987
123
another attempts at getting part b)

kx = -bv

kx = -2(km)^1/2(omegaA)

x = (-2(km)^1/2(omegaA))/k = -.100

didn't want to plug it into masteringphysics since I have only one try left, is this correct? (or am I getting warmer?)
 
  • #4
987
123
another attempts

x = Ae^((-b/2m)t)cos(omega't+phi)

omega' = (k/m-b^2/4m^2) = (k/m-k/m) = 0

cos(0) = 1

x = Ae^((-2(km)^1/2/2m).320)(1) = .0080

not sure about this one since this is for damped oscillations?
 
  • #5
alphysicist
Homework Helper
2,238
1
Hi clope023,

I believe your attempt at part b in the original post was close. However, if you'll notice that the equation you use does not match the equation you list in your relevant equations section; it's missing the phase constant phi.

To see what's happening, with your equation, you're using

x = A cos( omega t)

We want the object at t=0 to be at the equilibrium position, but at t=0 the equation you used indicates that object is at x=A. (So it looks like what you found was the position of the object 0.32 seconds after the object reached the highest point.)

To match the functional forms with the behavior at t=0, there are 4 special cases that are good to know:

x=A cos(omega t) : at t=0, object is at x=A
x=-A cos(omega t) : at t=0, object is at x=-A
x=A sin(omega t) : at t=0, object is at equilibrium position (x=0) and moving towards positive x
x=-A sin(omega t) : at t=0 object is at equilibrium position (x=0) and moving towards negative x

(These ideas come from looking at plots of all four functions.) If it's not one of those special cases you'll need to use a function with a phase shift phi that gives the function the proper behavior with time.

-----------------------------------------

About part c: What if they had asked for the force that the spring exerts on the object when the object is at the equilibrium position? We are using x=0 to be the equilibrium position, but the spring force there is not zero.
 
  • #6
25
0
part b:
if using x = A cos(omega*t), we must first determine when it has passed the equilibrium position, moving downward.
x = A cos(omega*t) = 0
dx/dt = -A*omega sin(omega*t) < 0
so omega*t = pi/2
then we want to find x = A cos(omega*(t + 0.320)) = A cos(pi/2 + omega*0.320) = -A sin(omega*0.320)

part c: when it is a distance 2.50×10−2m below the equilibrium position (no matter moving upward of downward), x = -2.50×10−2m
But f = kx is NOT the force F exerted by spring, it is the net force. 2 forces act on the object. At x = -2.50×10−2m, the amount of stretch of spring is x1 + x, where kx1 = mg, so the required force is k(x1 + x) = mg + kx, with upward direction.
 

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