Two masses which are connected each other with a spring

In summary, the problem involves two blocks with mass m on an inclined plane with angle \theta, connected by a spring with spring constant k and initial length L+x_0. The coefficient of kinetic friction for both blocks is \mu _k. The goal is to find the displacement of the blocks along the incline as a function of time, assuming that the spring is still stretched. This can be solved using two time-dependent equations and the general solution of Hooke's Law. The force of friction acts in the opposite direction and the displacement of both blocks affects the spring force.
  • #1
Diff.Ed
5
0
Homework Statement

NBRgW.png

Two blocks both of mass [itex]m[/itex] are placed on an inclined plane making an angle [itex]\theta[/itex] with the ground. The two masses are connected with a spring of spring constant [itex]k[/itex]. The coefficient of kinetic friction is [itex]\mu _k[/itex] for both blocks. Assume that the spring is initially stretched to a length [itex]L+x_0[/itex] where [itex]L[/itex] is its equilibrium length when it is at rest on a flat surface. Find the displacement of the two blocks ALONG THE INCLINE as a function of time,t, assuming that at that particular time the spring is still stretched. (Hint : Call the displacement of the leading block along the inclined plane [itex]x_1[/itex]. Write the displacement of the trailing block in terms of [itex]x_1[/itex] and stretch in the spring, [itex]x[/itex]. You will have two time-dependent equations in two unknowns.)

Related Equations
F=-kx (Hooke's Law), differential equations

The attempt at a solution
XN13k.png

[itex]m\frac{d^2x_1}{dt^2}=-k(x_1+x_0)-mg\sin\theta+\mu _k mg\cos\theta[/itex]​
and general solution of this diff. equation
[itex]x_1 = A\sin(\sqrt{k/m}t) + B\cos(\sqrt{k/m}t) - \frac{kx_0+mg\sin\theta+\mu _k mg\cos\theta}{k}[/itex]
XN13k.png

[itex]m\frac{d^2x_2}{dt^2}=k(x_1+x_0)-mg\sin\theta+\mu _k mg\cos\theta[/itex]​

I couldn't continue anymore. Could you help me ?
 
Last edited:
Physics news on Phys.org
  • #2
Draw the direction of motion of both bodies: The force of friction acts in the opposite direction. Note that the displacement of both bodies influences the length of the spring, so both x1 and x2 influences the spring force.

ehild
 

1. What is a spring constant?

A spring constant is a measure of the stiffness of a spring. It is represented by the letter "k" and is measured in units of force per unit of length (N/m or lbs/in).

2. How does a spring affect the motion of two connected masses?

A spring can either compress or stretch when two masses connected by the spring are moved. This causes a force to be applied to both masses, resulting in oscillatory motion between the two masses.

3. What factors affect the oscillation of the two masses connected by a spring?

The oscillation of the two masses is affected by the mass of each object, the spring constant, and the initial displacement or force applied to the system.

4. How do I calculate the period of oscillation for the two masses connected by a spring?

The period of oscillation can be calculated using the formula T = 2π√(m/k), where T is the period in seconds, m is the combined mass of the two objects, and k is the spring constant.

5. Can the spring constant be changed?

Yes, the spring constant can be changed by altering the material, length, or thickness of the spring. It can also be changed by changing the number of springs connected in series or parallel.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
461
  • Introductory Physics Homework Help
Replies
2
Views
617
  • Introductory Physics Homework Help
Replies
8
Views
323
  • Introductory Physics Homework Help
Replies
4
Views
730
  • Introductory Physics Homework Help
Replies
1
Views
798
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
29
Views
908
  • Introductory Physics Homework Help
Replies
16
Views
387
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
984
Back
Top