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Consider a mass $m$ at position $x(t)$ on a rough horizontal table attached to the origin by a spring with constant $k$ (restoring force $−kx$) and with a dry friction force $f$

\begin{equation} f= \begin{cases} F\quad \quad \quad \quad \quad \quad \,\,\,\, x<0\\ -F \leq f \leq F \quad \quad x = 0\\ -F \quad \quad \quad \quad \quad \quad x>0 \end{cases} \end{equation}

What is the range of $x$ where the mass can rest? Show that if the mass moves, the maximum excursion decreases by $2F/k$ per half cycle. Discuss the motion.

Up until now i have only dealt with SHM with no friction so I am a bit lost. Any help is appreciated.

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  • $\begingroup$ Your friction term is not written correctly, you have two cases for $x>0$. $\endgroup$
    – Ian
    Commented Nov 28, 2015 at 15:51
  • $\begingroup$ See this. $\endgroup$
    – A.S.
    Commented Dec 7, 2015 at 12:40

1 Answer 1

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Some Hints

The governing differential equation according to Newton's second law of motion will be

$$\begin{cases} - kx - F = m\ddot x & \dot x \gt 0 \\ - kx + F = m\ddot x & \dot x \lt 0 \end{cases}$$

If $\dot x=0$ then we should investigate that whether the particle moves or not due to the laws of static and kinetic friction. Rearranging the terms will give

$$\begin{cases} \ddot{x} + \omega^2 x = - \frac{F}{k} & \dot x \gt 0 \\ \ddot{x} + \omega^2 x = \frac{F}{k} & \dot x \lt 0 \end{cases}$$

where $\omega = \sqrt{\frac{k}{m}}$ is the natural frequency. The general solution to this ODE is

$$\begin{cases} x(t) = A \cos (\omega t) + B \sin (\omega t) - \frac{F}{k} & \dot x \gt 0 \\ x(t) = A \cos (\omega t) + B \sin (\omega t) + \frac{F}{k} & \dot x \lt 0 \end{cases}$$

Try to discuss the problem based on initial conditions.


Reference

Here is a nice article that you can refer to it. It explains everything from A to Z. You can take a look at part IV. There is a little mistake in equations 22a, 22b and 23a, 23b. The $\dot{x}$ should be written instead of $\ddot{x}$ for the signs. The text mentions this issue by words right above equations 22a, 22b so it seems to be a simple typo.

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  • $\begingroup$ Could you provide a complete solution? $\endgroup$
    – Robert S
    Commented Dec 7, 2015 at 0:21
  • $\begingroup$ @RSparkes: I added more hints. I want you to solve it by your self. :) $\endgroup$ Commented Dec 7, 2015 at 9:18
  • $\begingroup$ You solution is incorrect as it's discontinuous. Also $sgn'=2\delta\ne 0$. $\endgroup$
    – A.S.
    Commented Dec 7, 2015 at 10:04
  • $\begingroup$ What does F and f signify, also what is a dry friction force? Also what is sgn? $\endgroup$
    – Robert S
    Commented Dec 7, 2015 at 10:42
  • $\begingroup$ @A.S.: Maybe you are right but I should think more. :) However, we are taking the derivative with respect to $t$ and not $\dot x$! Physically, we have a force that its direction depends on the sign of velocity. Please feel free to edit my answer if you are sure about something or post a new answer that we can learn from it. :) $\endgroup$ Commented Dec 7, 2015 at 11:58

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