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I must determine if the system (a) is in SHM. I must arrive to the conclusion that there is a linear restoring force such that $a=-w^2x$.

The solution's manual affirms that according to Newton's third law, $k_1x_1=k_2x_2$. I tried to draw the free-body diagrams for the mass and the second spring, and another one for the mass only. However, I do not see that $k_1x_1=k_2x_2$.

Can anyone explain?

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  • $\begingroup$ the question on the title does not have anything to do with what you are asking here.... $\endgroup$
    – quarkleptonboson
    Dec 6, 2014 at 15:26
  • $\begingroup$ I think I am having trouble with the FBD of the mass. What forces should I have? Gravity, normal force, $k_2x_2$? How does the spring 1 have impact? $\endgroup$
    – yolo123
    Dec 6, 2014 at 15:37

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Take a look at the junction of the two springs. If you displace the block and hold it there, the junction will also be at rest.

At the junction, two forces are pulling it, $k_1 x_1$ to the left and $k_2 x_2$ to the right.

Since the junction is at rest, the two forces must balance. Therefore, $$k_1 x_1 = k_2 x_2$$

EDIT 1:

Another calculus-based derivation which I feel is more fundamental:

For a given displacement of the block, the springs will distribute the total elongation between themselves in such a way that the total potential energy of the system is minimum. $$P.E=\frac 12 k_1 {x_1}^2+\frac 12 k_2 {x_2}^2$$

Since the potential energy is to be minimum for a given displacement of the block, we take its derivative with respect to any one of the elongations (say $x_1$) and equate it to zero. (You can always check whether it is actually a minima by doing the second derivative test). $$\frac {d(P.E)}{d{x_1}}=\frac 12 k_1 (2x_1)\frac {dx_1}{dx_1}+\frac 12 k_2 (2x_2)\frac {dx_2}{dx_1}=0$$ $$\implies k_1 x_1 + k_2 x_2 \frac {dx_2}{dx_1} = 0$$

As $x_1 + x_2=x \implies x_2=x-x_1 $ (where $x$ is the given displacement of the block),$$k_1 x_1 + k_2 x_2 \frac {d(x-x_1)}{dx_1} = 0$$ $$\quad \implies k_1 x_1 + k_2 x_2 (0-\frac {dx_1}{dx_1}) = 0\quad $$ $$\quad \implies k_1 x_1 + k_2 x_2 (-1) = 0\quad $$ $$ \implies \boxed {k_1 x_1 = k_2 x_2} $$

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  • $\begingroup$ What about the mass m, what force acts on it? $\endgroup$
    – yolo123
    Dec 6, 2014 at 15:39
  • $\begingroup$ the resulting force is the effective elastic force of the two springs combined. once you have calculated the effective spring constant of that combination, the force is simply $-k_{eff} x $. where x is the displacement from equilibrium position. $\endgroup$
    – quarkleptonboson
    Dec 7, 2014 at 9:40
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simple. the box is confined in space by two springs. there are only two forces acting on the box, $F_1$ and $F_2$, exerted by those two springs.

Well, the two forces must balance each other out by Newton's third law, or the law of action-reaction pairs. so we get

$$ F_1 = -F_2 = -k_1 x_1 = k_2 (-x_2) --> k_1 x_1 = k_2 x_2 $$

(note that $x_1$ and $x_2$ increase going to the right, like usual x-coordinates.

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  • $\begingroup$ What about the mass m, what force acts on it? $\endgroup$
    – yolo123
    Dec 6, 2014 at 15:39

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