0
$\begingroup$

This is from Hobson, Riley, Bence Mathematical Methods problem 9.1

The question is: enter image description here

There is a hint given: enter image description here

If someone could just help me understand the physics part, I can try to do the calculations alone. What exactly is shown in this image? Are these hanging masses attached by springs if so why is the motion perpendicular to each other and not towards each other? What are these formulas above representing? I see a Force on the left but I don't recognize the right side. Why is no equation of energy represented like $\frac{mgx_1^2}{2l}+\frac{mgx_2^2}{2l}+\frac{1}{2}k(x_2-x_1)^2$


Does it look like this? (Note: This is a random photo from the internet not from the text.) enter image description here

$\endgroup$
  • $\begingroup$ Top diagram masses m are moving exactly out of phase and M is not moving. Middle diagram has all three masses oscillating in phase with the same amplitude. Bottom diagram the masses m are oscillating in phase with one another with M exactly out of phase with them. The amplitudes of masses m and M are different as shown in the diagram. $\endgroup$ – Farcher Jun 12 '17 at 12:08
1
$\begingroup$

Edited

What exactly is shown in this image? Are these hanging masses attached by springs

No, by stiff rods with a rotational spring.

if so why is the motion perpendicular to each other and not towards each other?

Because "they swing perpendicularly to the horizontal line containing their point of suspension." according to your transcript. Meanig they all swing in parallel planes.

The image is misleading in that it doesn't picture their vertical support beams. Forget the lines between the masses, they are just there to visualize the modes.

Here is the problem without the coupling springs of stiffness d pictured (Excuse the poor ascii art), viewed so the masses swing out of the plane:

=O=====O=====O===  <-- Horizontal line with 3 points of suspension
 |     |     |
 |     |     |
 |     |     |    <-- 3 rods, assuming same length 'c'
 |     |     |
 m     M     m    <-- 3 masses m, M, m

Now the coupling springs are additional springs between the masses, shown here:

Here is the view with the line of suspension sticking out of the screen, with the coupling springs:

        0   <--  Line of suspension            
       /|\
      / | \
     /  |  \  
    /   |   \   
   m^^^^M^^^^m  <-- three masses with a coupling spring 'd' between them
                    M is shown vertical and in the middle, but is swinging like the others

   <----|        x1
        |---->   x3
        |        (x2 not shown, is zero in this position)

Note that using the small displacement hypothesis, $x \approx c \times \sin(theta)$

What are these formulas above representing? I see a Force on the left but I don't recognize the right side.

They are the Newton's law of motion expressed for each particle, where $x$ is the (small) displacement of the mass.

Left hand side is actually $mass \times acceleration$

Right hand side is the sum of forces on each mass: the term in $c$ is their respective weight contribution ($F=m \times c \times sin(theta) \approx m \times \c \times x$), the terms in $d$ are the coupling spring's contributions.

Why is no equation of energy represented like $\frac{mgx_1^2}{2l}+\frac{mgx_2^2}{2l}+\frac{1}{2}k(x_2−x_1)^2$

The author did not try to represent the energy of the system or of each mass at all - but you could try.

It would not quite be $\frac{mgx_1^2}{2l}+\frac{mgx_2^2}{2l}+\frac{1}{2}k(x_2−x_1)^2$ for each mass, however, because there are more than one spring connected and their stiffness are not $k$.

But I'll leave you to find it, now you have a clearer picture. Don't hesitate to ask for more info.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I'm still a bit confused. Is c a spring constant and then cm mean c*m? If so what force is cmx? What does the pendulum look like? Is it formed like the photo added? You say "vertical springs" are they bouncing up and down and then how are they coupled? $\endgroup$ – user5389726598465 Jun 12 '17 at 12:24
  • $\begingroup$ The five springs in your asci art don't look like a pendulum, they look like they are bouncing vertically unattached with two springs on the left and right and one in the middle. The hint photo looks like the masses are moving in the horizontal plane "as viewed from above" $\endgroup$ – user5389726598465 Jun 12 '17 at 12:34
  • $\begingroup$ 1. 'c' is a constant but not the typical stiffness one, it must be multiplied by mass. So cm (yes multiply) is the stiffness 'k'. 2. Cmx is the force like k(x-x0) where k=cm and x0=0. 3. Yeah I'll re-write my answer, I modelled spring masses instead of pendulums, my bad $\endgroup$ – MrBrushy Jun 12 '17 at 13:08
  • $\begingroup$ Not in your question, but can we assume small displacement? Otherwise the eq. will not hold $\endgroup$ – MrBrushy Jun 12 '17 at 13:09
  • $\begingroup$ Sure, assume small displacement. $\endgroup$ – user5389726598465 Jun 12 '17 at 13:28

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.