Damped Harmonic Motion on a Spring

In summary, the problem involves investigating damped harmonic motion of a block on a horizontal table attached to a spring. The coefficient of friction, equilibrium position, and amplitude of the block are given. Using Newton's 2nd law, the equation for acceleration is found to be (-kx+mu*mg)/m. The position of the block is given by x(t)=x'*cos(omega*t) for the first half cycle and x(t)=x''*cos(omega*t) for the second half cycle. The problem asks for a graph of x(t)/x0 vs t/T and it is noted that something interesting happens at the end of the 5th half cycle, where the motion may become flat as the block
  • #1
Johnny0290
9
0

Homework Statement



In this problem we will investigate a particular example of damped harmonic motion. A block of mass m rests on a horizontal table and is attached to a spring of force constant k. The coefficient of friction between the block and the table is mu. For this problem we will assume that the coefficients of kinetic and static friction are equal. Let the equilibrium position of the mass be x = 0. The mass is moved to the position x = +A, stretching the spring, and then released.

2.1 Apply Newton's 2nd law to the block to obtain an equation for its acceleration for the first half cycle of its motion, i.e. the part of its motion where it moves from x = +A to x < 0 and (momentarily) stops. Show that the resulting equation can be written d2x'/dt2 = -omega^2 * x', where x' = x - x0 and x0 = mu*m*g/k. Write the expression for position of the block, x(t), for the first half cycle (be sure to express omega, the angular frequency, in terms of the constants given in the problem statement). What is the smallest value of x that the mass reaches at the end of this first half cycle?


2.2 Repeat the above for the second half cycle, i.e. wherein the block moves from its maximum negative position to its (new) maximum positive position. First show that the differential equation for the block's acceleration can be written d2x''/dt2 = -omega^2*x'' where this time x'' = x + x0. Next, match the amplitude for the beginning of this half cycle with the amplitude at the end of the last one. Write the expression for the position of the block, x(t), for the second half cycle.


2.3 Make a graph of the motion of the block for the first 5 half cycles of the motion in the case where A = 10.5*x0. Plot the position of the block normalized to x0 as a function of the fractional period, T = 2*Pi/omega (i.e. plot x(t)/x0 vs t/T). Attach the graph to your email.


2.4 Something interesting happens at the end of the 5th half cycle - what changes in the physical situation and what is the motion after this half cycle?

Homework Equations



F=ma
F=-kx
F_friction=mu*m*g

The Attempt at a Solution



I was able to get answers for 2.1 and 2.2

Acceleration
F_spring-F_friction=m*a
-k*x+mu*m*g=m*a
a=(-kx+mu*mg)/m
This works for 2.1 and 2.2 and I am able to simplify it to -omega^2*x' and -omega^2*x''

Position
x(t)=x'*cos*(omega*t) for 2.1
x(t)=x''*cos*(omega*t) for 2.2

x' and x'' seemed like the amplitudes at each of those intervals so I plugged them into the position equation for general harmonic motion.

Now I'm working on 2.3 and confused about graphing this.
I'm not sure why the question wants x(t)/x0 vs t/T.
Won't t/T give me a unit-less x-axis since time and radians cancel out?
The concept of x' and x'' confuses me. Does anyone care to explain?

2.4
I haven't graphed it but I'm guessing the graph starts to go flat or something like that as the mass begins to reach equilibrium.
 
Physics news on Phys.org
  • #2
I'm not sure what the physical situation is supposed to be and what the motion after the half cycle is.
 

What is damped harmonic motion on a spring?

Damped harmonic motion on a spring is a type of oscillatory motion in which a mass attached to a spring experiences a resistive force that reduces its amplitude over time. This is known as damping, and it can occur due to friction or air resistance.

What is the equation for damped harmonic motion?

The equation for damped harmonic motion is x(t) = A * e^(-bt/2m) * cos(ωt + φ), where A is the initial amplitude, b is the damping coefficient, m is the mass, ω is the angular frequency, and φ is the phase angle.

How is damped harmonic motion different from simple harmonic motion?

Damped harmonic motion differs from simple harmonic motion in that it experiences a resistive force that reduces its amplitude over time. In simple harmonic motion, there is no resistive force and the amplitude remains constant.

How is the damping coefficient related to the amount of damping in damped harmonic motion?

The damping coefficient, b, is directly proportional to the amount of damping in damped harmonic motion. This means that a larger value of b will result in a greater reduction in amplitude and a faster decay of the oscillations.

What is critical damping in damped harmonic motion?

Critical damping in damped harmonic motion occurs when the damping coefficient is equal to the critical damping constant, which is equal to 2√(km) where k is the spring constant and m is the mass. At this point, the system will return to equilibrium without any oscillation.

Similar threads

  • Introductory Physics Homework Help
2
Replies
51
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
395
  • Introductory Physics Homework Help
Replies
17
Views
372
  • Introductory Physics Homework Help
Replies
5
Views
834
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
918
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top