What Is the Energy of a Viscously Damped Spring-Mass System?

In summary: I just know their program is wrong, they have the mass at .45 instead of 7.5, but it looks like it's just a copy paste of their previous program.
  • #1
jchojnac
39
0

Homework Statement


Use your program for the viscously damped spring to solve this problem. Make sure you change the parameters as specified below.

A mass m=7.5 kg is attached to the end of a spring with a spring constant of k=18.3 N/m. The mass moves through a viscous damping medium with a damping constant b=1.8 kg/s giving a velocity dependent damping force Fdamp= -bv.

The motion occurs in zero gravity so set the force of gravity to ZERO in your program. Also set the equilibrium position L0=0. The mass is initially motionless and displaced from equilibrium by a distance yinitial=0.2 m.

What is the energy of the spring-mass system at the initial position of the mass? (the spring-mass system does not include the damping medium)

What is the energy of the spring-mass system when the mass first passes through the equilibrium position? (you may wish to include a logical test to help you find when this occurs)


Homework Equations


1/2mv^2
mgy


The Attempt at a Solution


just tried m*g for the first one so far.
 
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  • #2
Without knowing what your program is, it's a little hard to know what you are struck on.
 
  • #3
from __future__ import division
from visual.graph import *

scene.y = 400 # moves animation window down 400 pixels to make room for graph# Constants
pi = 3.14159
L0 = .08 # equilibrium length of spring (not stretched)
g = 0 # gravitational acceleration set to zero (not on earth)
k = 14 # insert the spring constant you found for the coiled spring

# Objects
# Ceiling to hang spring from.
# block to act as mass.
# Spring is represented by a cylinder.
scene.center = vector(0,-.1,0) # you may want to adjust this to improve display

ceiling = box(pos=vector(0,0,0), size=(.3,0.005,0.005)) # make the contact pointthe origin
block=box(pos=vector(0,-0.1,0), size=(.02,0.02,0.02), color=color.yellow)

# Using the positions of the block and ceiling set the cylinder to stretch from the ceiling to the block
spring = cylinder(pos=ceiling.pos, axis=block.pos, radius=.005)

# Initial values
block.m = 0.45 # insert the measured mass from coiled spring experiment.
block.v = vector(0,0,0) # the vector velocity assuming the block is initially stationary
block.p= block.m * block.v

block.pos=vector(0,-L0-0.05,0) # initial position of block 0.05m from equilibrium

# Setting the timestep and zeroing the cumulative time
deltat = .0001 # you should decrease this later to test if it is small enough
t = 0
W = 0
displacement=0
Kgraph = gcurve(color=color.cyan)
Ugraph = gcurve(color=color.yellow)
KplusUgraph = gcurve(color=color.red)
Wgraph = gcurve(color=color.green)
# Loop for repetitive calculations
scene.autoscale=0
while t < 4:

Fnet= -(((block.pos-vector(0,-L0,0))*k))-((block.p/block.m)*.2) #INSERT the force of the spring on the block
displacement=(mag(block.p)/block.m)*deltat
block.p= block.p+Fnet*deltat # updates the momentum
block.pos= block.pos+block.p/block.m*deltat # updates the position
spring.axis = block.pos #updates the spring axis so it stays on the block
t=t+deltat
pmag = mag(block.p) # or whatever you've called the block's momentum

K = (pmag**2)*.5/block.m #COMPLETE this for the kinetic energy of the
block
U = ((mag(block.pos)-L0)**2)*.5*k #COMPLETE this for the potential energy of the block-spring system (note no gravity)
W = W - displacement*(mag(block.p)/block.m)*.2
Kgraph.plot(pos=(t,K))
Ugraph.plot(pos=(t,U))
KplusUgraph.plot(pos=(t,K+U))
Wgraph.plot(pos=(t,W))
 
  • #4
this is the program need to solve this problem, but i don't know what # need to be changed to solve this problem

if someone could help me that would be great
 
  • #5
hi the energy of the spring-mass system at the initial position of the mass will be
the potential energy only which is
1/2 * k * x^2

and then plug the number in
.5 * 18.3 * (.2)^2 = .366
 
  • #6
how would you solve their 2nd question?!

"What is the energy of the spring-mass system when the mass first passes through the equilibrium position? (you may wish to include a logical test to help you find when this occurs)"
 
  • #7
Well I'd assume that Since Ef = Ei + Wsurr,
And they gave us how to solve the Fdamp = -bv.
So would it be something like Ef = Ei + (Fdamp * x)? Just a guess, I'm in the same mess.
 

Related to What Is the Energy of a Viscously Damped Spring-Mass System?

What is a spring mass system?

A spring mass system is a physical system where a mass is attached to a spring, allowing the mass to oscillate back and forth when the spring is stretched or compressed.

What is the equation for the energy of a spring mass system?

The equation for the energy of a spring mass system is E = 1/2 * k * x^2, where k is the spring constant and x is the displacement of the mass from its equilibrium position.

How does the energy of a spring mass system change as the mass oscillates?

As the mass oscillates, the potential energy of the system is constantly changing into kinetic energy and back. However, the total energy of the system remains constant.

What factors affect the energy of a spring mass system?

The energy of a spring mass system is affected by the spring constant, the mass of the object, and the amplitude of the oscillation.

What is the relationship between the energy of a spring mass system and its frequency of oscillation?

The energy of a spring mass system is directly proportional to the square of its frequency of oscillation. This means that as the frequency increases, the energy also increases.

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