SImple Harmonic Oscillator under constant friction force

[henryc09]

Homework Statement

There is a mass attached to two springs on a table. Coefficients of static and sliding friction between the mass and table are equal with the value $$\mu$$.

The particle is released at time t=0 with a positive displacement x₀ from equilibrium. Given that 2kx₀ > $$\mu$$mg write down the equation of motion as long as it remains moving. Verify it's satisfied by x(t)=Acos(wt) + Bsin(wt) + C and find the constants A,B and C for the data given.

Find the time t₁ and position x₁ at which the particle next comes to rest.

Homework Equations

The Attempt at a Solution

I think I can do the verifying part, and am able to get B=0 after differentiating to find v which is 0 when t=0. Then A+C = x₀ I think. The equation of motion is
ma=-2kx+ $$\mu$$mg i guess? Really don't know where to go from here. I worked out earlier in the absence of friction that the angular frequency w=sqrt(2k/m), is this still the same? Any ideas what to do next?

 

[anathema]

Hello, thank you for your post. It's great to see you working through this problem and asking for help when you need it. Let's see if I can offer some guidance.

First, your equation of motion is correct: ma = -2kx + μmg. This accounts for the two springs and the force of friction acting on the mass.

Next, you are correct that the angular frequency w is still the same in the presence of friction. This is because the spring constants and mass do not change, so the natural frequency of the system remains the same.

To find the constants A, B, and C, you will need to use initial conditions. At t=0, the particle has a positive displacement x0 from equilibrium. This means that x(0) = x0. You can also use the initial velocity, v(0), to find the constant B.

To find the time t1 and position x1 at which the particle next comes to rest, you will need to use the fact that when the particle comes to rest, its velocity v(t) = 0. You can use this to find the time t1, and then plug that value into the equation for position x(t) to find x1.

I hope this helps. Keep up the good work!

 

[kerimek]

I would approach this problem by first breaking it down into smaller parts and understanding the concepts involved.

First, let's look at the equation of motion for a simple harmonic oscillator without any friction: ma = -kx. This equation tells us that the acceleration of the mass is directly proportional to its displacement from equilibrium, with a constant of proportionality of -k/m. The angular frequency, w, can be calculated as w = sqrt(k/m).

Now, let's introduce the friction force into the equation. As stated in the problem, the friction force is equal to the coefficient of friction, \mu, multiplied by the mass of the object and the acceleration due to gravity, mg. Therefore, the equation of motion becomes ma = -kx - \mu mg.

Next, we need to find the constants A, B, and C for the given data. As you correctly stated, B = 0, and A + C = x0. To find the value of A and C, we can use the initial conditions given in the problem. At t=0, the displacement of the mass is x0, and its velocity is 0. Therefore, we can set up the equations x0 = A + C and 0 = -Aw + Bw, where w is the angular frequency. Solving these equations, we get A = x0 and C = 0.

Now, to find the time t1 and position x1 at which the particle next comes to rest, we need to set the velocity of the mass to 0. From the equation of motion, we know that the velocity is 0 when -kx - \mu mg = 0. Solving for x, we get x = - \mu mg/k. This is the position at which the particle comes to rest. To find the time, we can use the equation of motion and set the acceleration to 0, which gives us t1 = w/2pi. Plugging in the values for w and solving for t1, we get t1 = sqrt(m/k)/2pi.

In summary, the equation of motion for a simple harmonic oscillator under constant friction force is ma = -kx - \mu mg, and the constants A, B, and C for the given data are A = x0, B = 0, and C = 0. The position x1 at which the