2
$\begingroup$

Apologies if math stackexchange is the wrong place to post this, I'm stuck on a differential equations/physics problem in a DE textbook.

An object of mass $m$ is attached to the midpoint of a light elastic string of natural length $6a$. When the ends of the string are fixed at the same level a distance $6a$ apart and the mass is allowed to hang in equilibrium, the length of the stretched string is $10a$. The mass is pulled down a small vertical distance from equilibrium and released.

Show that for small oscillations, the period of the resulting motion is $T = \frac{20 \pi}{7} \sqrt{\frac{a}{g}}$, where I guess $g$ is the acceleration due to gravity.

I'm in a rather sad state. I've never taken physics in my life, and I'm trying to learn. But I don't really get what the question is asking, it seems too vague. I think we can relate this problem to a problem about springs: it seems like we could draw an imaginary vertical line from the object up to the imaginary horizontal line connecting the endpoints of the string (the horizontal line is where the string would be if there were no object attached to it and pulling it down), and that imaginary vertical line is basically a spring, subject to Hooke's law and whatnot.

In that case, the "spring" would have equilibrium length $\sqrt{(5a)^2 - (3a)^2} = 4a$ when the object of mass $m$ is attached to it. I know there is an equation governing the motion of the object of mass $m$: $$y'' + \frac{c}{m} y' + \frac{k}{m}y = F(t)$$ where $c$ is a "damping constant" (don't get what that is), $k$ is the spring constant which can be obtained from Hooke's law (I think I know how to find it), and $F(t)$ is some function which describes an external force affecting the spring.

What does "small oscillations" mean anyway? Does that mean that $y$ begins with some small initial velocity? Or that $F(t)$ is some function which takes on small values? So confused right meow.

$\endgroup$
  • 1
    $\begingroup$ Did you draw a diagram? That should make things much clearer. Note that this system is effectively a mass balanced by two springs both at an angle to the vertical. Small oscillations here means that the angle doesn't change appreciably with the motion, so you can stay in the simple harmonic motion scenario (I think you're supposed to neglect damping largely because the question doesn't mention it). Also, physics.stackexchange.com seems like the natural place for this question. $\endgroup$ – stochasticboy321 Dec 3 '15 at 4:16
1
$\begingroup$

This is a very easy problem. In this case, the oscillator does not have damping, and it is not driven. So the Differential equation will be of the form

$$y''=-\omega^2 y$$

And the time period will be $$T={2\pi\over\omega}$$

Small oscillations means that Hooke's Law $F=−kx$ is valid. In case of pendulums, it means that the angle is small and you can take $\sinθ≈θ$.

In this case to find $\omega$, you should draw a diagram, write all the forces, and apply Newton's second law, $F=ma$. I highly recommend you to watch Walter Lewin's lectures for understanding this.

EDIT: General meaning of small oscillation

The harmonic oscillator has a potential that is quadratic in the position and has a minimum at $x=0$. Any system which is in stable equilibrium has a minimum of potential energy at that point. So it may be approximated by a quadratic potential if the oscillations are not too large. Hence, any system in local minimum of potential energy behaves approximately like a harmonic oscillator.

E.g. see this diagram, Black is the actual potential while red is the quadratic approximation. So this system will behave approximately like the harmonic oscillator.

enter image description here

So general meaning of small oscillation is that the amplitude is not large and you can take the system to be a harmonic oscillator.

$\endgroup$
  • $\begingroup$ Thanks for answering. But I still don't get it. What does this have to do with a pendulum? $\endgroup$ – D_S Dec 3 '15 at 4:28
  • $\begingroup$ @D_S The question doesnt have anything to do with pendulums, but the case of the pendulum is similar and in that case there is another interpretation of small oscillations. In fact there is a more general interpretation of small oscillation which I am going to edit into my answer. $\endgroup$ – Kartik Dec 3 '15 at 7:17

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.