How Does Friction Affect Oscillations in a Spring-Mass System?

[toastie]

Homework Statement

The block oscillates about equilibrium for the spring. A weak frictional force of constant magnitude 2.7 N causes the oscillations to diminish slowly. The block oscillates many times and eventually comes to rest. First, show that the decrease in amplitude is the same for each cycle of oscillation. now, suppose the initial displacement from quilibrium is 0.57 m. How many cycles will occur before the block comes to rest?
spring constant k = 51.3 N/cm

Homework Equations

mx"w^2 + 0.57 - 2.7 = 0

x" = the second derivative of x

The Attempt at a Solution

I keep trying find an equation to solve this problem, but I keep getting the above equation which I can't solve for because I have 2 unknowns.

Am I taking the right approch to this problem?

Thank you in advance!

 

[Hootenanny]

mx"w^2 + 0.57 - 2.7 = 0

I'm not quite sure how you've constructed that equation, but as far as I can tell it is incorrect.

Can you use Newton's second law to create an equation of motion?

 

[toastie]

I have done an energy and force anaylsis. They both give the following equation: mx" - kx - constant magnitude = 0

 

[Hootenanny]

I have done an energy and force anaylsis. They both give the following equation: mx" - kx - constant magnitude = 0

So where does the w come from and where does the k disappear to in your expression?

 

[toastie]

I mistyped the equation. Sorry for the confusion about equation! The equation I just typed is the one that I keep getting.

 

[Hootenanny]

I mistyped the equation. Sorry for the confusion about equation! The equation I just typed is the one that I keep getting.

mx" - kx - constant magnitude = 0

Good. So, this is a standard ODE, do you know how to solve it?

 

[toastie]

I don't know how to solve this problem without the mass.

 

[Hootenanny]

I don't know how to solve this problem without the mass.

You don't need to explicitly know the mass - you are not asked for a numerical answer. Simply solve the ODE to obtain the required function.

 

[toastie]

What is ODE an abreviation for?

 

[Hootenanny]

What is ODE an abreviation for?

Ordinary Differential Equation, which I assume that you have studied already?

 

[toastie]

okay i realize what ode stands for. I have tried that and I get x(t) = A*cos(angular velocity*t)*e^(-(beta)t). From here we are not really sure where to go. We know that we need to find the number of cycles but beyond that we are confused.

 

[Hootenanny]

okay i realize what ode stands for. I have tried that and I get x(t) = A*cos(angular velocity*t)*e^(-(beta)t). From here we are not really sure where to go. We know that we need to find the number of cycles but beyond that we are confused.

Good, the exponential decay implies that the amplitude decays at a constant rate.

Now, you want to find the time when the velocity is zero.

 

[diazona]

Actually, x(t) = A\cos(\omega t) e^{-\beta t} isn't a solution to the equation

m x''(t) - k x(t) - C = 0

And anyway I don't think that's the proper equation for this situation. For one thing, the solution I get doesn't oscillate at all. Also, the direction of the constant force C switches back and forth every time the block reverses direction, and the equation as given doesn't account for that.

I don't mean to be a pain, but unless I'm missing something really stupidly simple, it does seem like something is wrong...

 

[toastie]

if the velocity = 0 then I am left solving x(t) = Ae^(-(beta)*t)
However, I do not know what beta is. Should beta be the spring constant?

 

[Hootenanny]

Actually, x(t) = A\cos(\omega t) e^{-\beta t} isn't a solution to the equation

m x''(t) - k x(t) - C = 0

You are indeed correct, good catch!

toastie: You should note that your expression requires an additional constant term in order to satisfy the ODE. Apologies for not catching it sooner :redfaced:

And anyway I don't think that's the proper equation for this situation. For one thing, the solution I get doesn't oscillate at all.

That is the correct ODE for this situation. What solution do you obtain? It is possible that the oscillator is over-damped in which case no oscillations would occur, but I haven't checked.

Also, the direction of the constant force C switches back and forth every time the block reverses direction, and the equation as given doesn't account for that.

But we do account for the change of direction of C, the form of the ODE means that the constant force C is always acting against the motion of the mass.

 

[diazona]

I get an exponential curve,
x(t) = c_1 e^{t\sqrt{k/m}} + c_2 e^{-t\sqrt{k/m}} - \frac{C}{k}
both from Mathematica and by hand. I have to admit, I didn't think about the possibility of the oscillator being overdamped - although the OP does say the block oscillates about equilibrium, which suggests an underdamped oscillator.

And I really don't think the equation as is accounts for the change in direction of the force. Here's my reasoning: I would expect a force that opposes the motion of the mass to be represented by a velocity-dependent term, as in the damped harmonic oscillator,
m x'' + b x' + k x = 0
Of course, that damping force depends not only on the direction of velocity, but on the magnitude of velocity - the faster the motion, the larger the force. In this case, the force is of constant magnitude, so we have to "normalize" that term, dividing out the magnitude of the velocity,
m x'' + b \frac{x'}{\vert x' \vert} + k x = 0
As a check, this is undefined when x' = 0, which is what I would expect from a force of constant magnitude that opposes the motion: when there is no motion, the force has a nonzero magnitude but no direction to act in, so it is "physically undefined." (\mathrm{sign}\,{x'} rather than x'/\vert x' \vert would avoid this singularity)

In the equation we've been dealing with,
m x'' - k x - C = 0
the term -C is of constant magnitude and constant sign, which represents a constant force of 2.7 \mathrm{N} acting in the negative x direction. That would just shift the equilibrium position of the spring; I can't see it accounting for a frictional force that always opposes the motion.

 

[Hootenanny]

Hmm, I must admit that I agree with you there. Before now I had just thought of this as 'standard' harmonic oscillator question, which of course it is not. The problem would also be straight forward if the damping term was velocity dependent, unfortunately, this is not the case, nor is the inhomogeneous term constant.

I'll also admit that the equation that we were originally working with is entirely incorrect as you point out. I should have read the OP fully.

So, let's take your proposed equation of motion (with the sign function) and see where we get:

x^{\prime\prime} + \frac{b}{m}\;\text{sign}\;x^\prime + \frac{k}{m}x = 0

Solving ODE's with the sign function is messy, so I propose that we explicitly split the EOM into the three distinct regions:

\left.\;\;\; \begin{aligned} x^{\prime\prime} - \frac{b}{m} + \frac{k}{m}x = 0 \;\;\;,\;\;\; x^\prime &gt; 0 \\<br /> x^{\prime\prime} + \frac{k}{m}x = 0 \;\;\;,\;\;\; x^\prime = 0 \\<br /> x^{\prime\prime} + \frac{b}{m} + \frac{k}{m}x = 0 \;\;\;,\;\;\; x^\prime &lt; 0 \end{array}\right\}

The equations in all three regions have the same homogeneous equation, with the corresponding indicial equation:

\lambda^2 + \frac{k}{m} = 0

Hence, the homogeneous solution can be written:

x = A^\prime\cos\left(\sqrt{\frac{k}{m}}t\right) + B\sin\left(\sqrt{\frac{k}{m}}t\right)

The particular integrals corresponding to regions one and three are simply additive constants (\mp \frac{b}{k}[/tex] respectively). For brevity, we recast the homogeneous solution exclusively in terms of cosine, with an arbitrary phase shift \phi. Hence, the solution in equation region may be written thus:<br /> <br /> \left.\;\;\; \begin{aligned} x = A\cos\left(\omega t -\phi\right)- \frac{b}{k} \;\;\;,\;\;\; x^\prime &amp;gt; 0 \\&lt;br /&gt; x = A\cos\left(\omega t -\phi\right) \;\;\;,\;\;\; x^\prime = 0 \\&lt;br /&gt; x = A\cos\left(\omega t -\phi\right) + \frac{b}{k} \;\;\;,\;\;\; x^\prime &amp;lt; 0 \end{array}\right\}.<br /> <br /> With \omega = \sqrt{k/m}.<br /> <br /> An important point to note here is that the above solutions do not represent a general solution for the displacement of the HO, rather they represent a family of solutions. The coefficient A is not constant throughout the whole motion, instead it is constant throughout each half oscillation.<br /> <br /> Before we process to assembling the piecewise smooth solution, can I get an agreement from <b>toastie</b> and <b>diazona</b> that we are all on the same page?