-1
$\begingroup$

Suppose we have the following setup, a mass on a spring attached to a wall, as shown in the diagram:

Image taken from google images

We are completely ignoring friction between the mass and the ground here, and we have an ideal spring too. The mass starts in the equilibrium position. Let's define a coordinate system such that the origin is the equilibrium position of the spring and the positive x axis to the right of the equilibrium point.

Let's now submerge this system in a fluid A, such that the motion of the mass is heavily over-damped (e.g molasses). We give the mass a kick to the right such that it has some initial velocity:

$$v = v_0$$ away from the wall.

Let's now submerge the system in a fluid B, so now the motion of the mass is critically damped. We give it the same initial velocity velocity away from the wall.

My question is, will the mass achieve a maximum displacement away from the equilibrium position (x=0) in fluid A or fluid B, or will the maximum displacements in each case be equal?

$\endgroup$
  • 4
    $\begingroup$ What do you think will happen and why? $\endgroup$ – sammy gerbil Jun 22 '17 at 3:42
  • $\begingroup$ My intuition tells me that the critically damped oscillator will achieve a higher maximum displacement, since the over damped oscillator is losing energy at a higher rate, and they start off with the same kinetic energy. The thing that was bugging me in my mind was that the critically damped mass will move more quickly through the fluid, though the overdamped mass will move more slowly, though over a longer period of time. $\endgroup$ – user154080 Jun 22 '17 at 4:44
  • $\begingroup$ As a result it's unclear to me which one will go further. $\endgroup$ – user154080 Jun 22 '17 at 4:49
1
$\begingroup$

This is an application of the Racetrack Principle : if horse B always runs faster than horse A, and if they start a race at the same place and time, horse B is bound to win.

Think of the two oscillators as being in a race. They start with the same speed from the same place. However, the heavily-damped oscillator (A) slows down more quickly. During every instant after the first, oscillator A is always moving slower than the critically-damped oscillator (B) and covers less distance. Oscillator A will stop first, because it is losing speed quicker. When it does so, oscillator B has travelled further in total, because during every instant it has travelled further. Furthermore, oscillator B continues moving away from the start position after A has stopped and turned back towards home.

$\endgroup$
  • $\begingroup$ That makes good sense. Just a point on the last paragraph, the form of over damped and critically damped motion is such that when released from equilibrium, they will not overshoot the equilibrium position. The critically damped oscillator will simply reach the origin faster. $\endgroup$ – user154080 Jun 22 '17 at 12:13
  • $\begingroup$ Yes, you are correct, it doesn't overshoot. I must have confused it with something else. I shall remove this paragraph. $\endgroup$ – sammy gerbil Jun 22 '17 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy