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I know that with the formula $T=2\pi\sqrt{\frac{m}{k}}$ the time period is not related to the amplitude. However, would amplitude matter if i do this experiment in real life. Would a greater amplitude result in more friction of some sort?

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  • $\begingroup$ It is an interesting question. When you pluck a guitar string real hard it begins with greater amplitude and diminishes but the frequency stays the same. $\endgroup$ – Bill Alsept Aug 16 '17 at 6:11
  • $\begingroup$ You're saying that the period of oscillation should remain constant ? $\endgroup$ – Hilkjh Aug 16 '17 at 7:41
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    $\begingroup$ @Bill Alsept it is not constant the frequency. If you pull it very much (but not that much as to break it) the string will vary slightly its frequency at the beginning. It is not much, but it is detectable. I have made the experiment with my guitars. I think in this case has to do with the fact that the string may go out of its linear regime for a little. $\endgroup$ – user171780 Apr 5 '18 at 23:40
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Ideally no. With "ideally" I mean that friction is proportional to velocity, the spring is ideal, and everything is independent of temperature and any other stuff out of Classical Mechanics.

In real life I bet for yes. This is because the formula $F_\text{friction} \propto -\dot{x}$ is a very simple model when temperature is constant, there are no turbulences in the fluid (or the surface), etc. In real life if you inject enough energy into the spring (this is equivalent to a very big initial amplitude) then dissipation will heat the surrounding thus changing the properties of the medium and thus varying not only the force of friction but also the properties of the spring (because it will heat also). In addition you can consider that the expression $F_{\text{spring}} = -k x$ is also an approximation, very good when $x$ is small but not to good for big values of $x$.

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$2$ $pi$ frequency is root of return force/(inertia- displacement). A real spring will also have nonlinear terms in return force. The return force can be expanded in powers of displacement such as $F = ax + bx.x + cx.x.x ..$.Where $x$ is the displacement and $a,b c$ etc are constants. $b < a, c < b$ etc. The nonlinear terms, that is those involving $b,c$ etc can be neglected for small oscillations. $2$ $pi$ frequency then takes the form as the root of $a/m$ . $m$ is the mass. This is constant. Note that as the displacement is made to increase by increasing the amplitude of oscillation terms involving $x$ square, $x$ cube etc become larger and cannot be neglected. The frequency will then depend on amplitude.

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  • $\begingroup$ So you're saying that spring constant (if large amplitude) would change therefore time period would change? $\endgroup$ – Hilkjh Aug 16 '17 at 8:48
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    $\begingroup$ The spring constant would not change with amp. However the return force could be taken as equal to kx +bkx*x+..... Where b<< 1. So the non linear terms like x square and x cube etc would start making a contribution to frequency giving T dependent on x for large enough x. $\endgroup$ – SAKhan Aug 16 '17 at 14:09
  • $\begingroup$ @SAKhan +1 it might be good to include your comment back into your answer, it's really central to your point. $\endgroup$ – uhoh May 7 '18 at 4:31
  • $\begingroup$ Done uhoh as you mentioned $\endgroup$ – SAKhan May 9 '18 at 18:56
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No, the amplitud does not affect to the period.

For a spring-mass system:

if you solve the differential equation $$m\ddot x +kx=0$$ you get a solution that looks like this $$x(t)=A_0 \cos (\omega t -\delta)$$

Where both the amplitude ($A_0$) and the phase angle ($\delta$) are in fact arbitrary constants which could be any real number, but the frequency $\omega $ is determined by the mass and the stiffness constant of the spring (or its equivalent in other mechanic system). And similarly to the damped oscillator.

At the moment this model works well enough to predict the results of all mechanics systems which performs small oscillations (like a pendulum, or a string), so an experiment in real life should behave as the theory says.

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  • $\begingroup$ The first equation is usually given for an ideal spring-mass system. When applied to a real world problem, most (good) authors will find a way to communicate in some way that it is an approximation, as you did by including "well enough" at the end of your answer. The answer is really "Yes, but not much" rather than "No." $\endgroup$ – uhoh May 7 '18 at 4:27

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