I have a spring of length $L$ when unstretched with one end fixed to the roof. At the lower end I place a mass $m$ and drop gently so it stretches by $x_1$ at equilibrium so that $mg=kx_1$. Now I place another mass $m$ on it and let it drop so that it oscillates. What will be the amplitude of this oscillation?

The equilibrium of the combined masses would be at $x_2$ below the initial equilibrium position and it can be shown that $x_1 = x_2= x$. By intution I feel it should oscillate with amplitude $x$ about the second equilibrium position. But when I try to work it out on energy conservation I get a different answer.

The elastic potential energy of the spring at the farthest position is $½k(x+y)^2$ where $y$ is the maximum position below initial equilibrium position. The changes in potential energy that are converted to elastic spring energy are $[mg(x_1+y)+ mgy]=mg(x_1+2y)$. Equating these and putting $mg= kx$ and solving the quadratic for $y$ gives $y=2.414x$ since $y$ cannot be negative.

I would be grateful if someone could point out where I am going wrong.

  • $\begingroup$ Are you considering that the first mass will also oscillate? $\endgroup$ – Mick Dec 22 '17 at 0:42
  • $\begingroup$ @Mick If I put the 2nd mass m on then the system consisting of 2m will oscillate.So yes.. $\endgroup$ – Chappy Dec 22 '17 at 2:34
  • $\begingroup$ Sorry, I imagined for some reason another spring and the second mass. My bad. $\endgroup$ – Mick Dec 22 '17 at 2:35
  • $\begingroup$ The spring has already stored some energy 1/2 k x^2 when the 2nd mass is dropped on it.Is this recovered in the form of potential energy when the spring bounces up? $\endgroup$ – Chappy Dec 22 '17 at 3:03

Consider, in your energy terms you will have elastic potential energy $(\frac12k\Delta x^2)$, gravitational potential energy $(mg\Delta h)$ and kinetic energy $(\frac12mv^2)$. At the limits of displacement the instantaneous velocity is 0 so kinetic energy is 0.

If the spring-mass system were oriented horizontally horizontal spring-mass

then $h$ stays at 0 so the gravitational potential energy is also 0. In this case your total energy will be $E_T=\frac12kx^2+mgh+\frac12mv^2=\frac12kx^2+0+0$. So at the limits of displacement the energy is simply $E=\frac12kx^2$ (and at the equilibrium point will be $E=\frac12mv^2$).

Then the spring-mass system would oscillate with amplitude $x$ about its equilibrium point with a period of $\sqrt{\frac km}$.

Now, when you orient the spring-mass system vertically then with a single mass $m$ the spring is initially stretched by an amount $x_1$ where $x_1=\frac{mg}k$, and with a second mass $m$ the spring is stretched by an amount $x_2$ where $x_2=\frac{2mg}k$.

If the spring-mass system is allowed to oscillate from a starting position of $x_1$, then again at the limits of displacement the instantaneous velocity will be 0 and so the kinetic energy will be 0, and your total energy will be $E_T=\frac12kx^2+mgh+\frac12mv^2=\frac12kx^2+mgh+0$.

$\therefore E_T =\frac12kx^2+mgh$

(and at the equilibrium point will be $E_T=mgh+\frac12mv^2$)

Now, when you consider the potential energy components you should arrive at the same solution independent of whether you set $h=0$ at the top of the oscillation, at the bottom of the oscillation or at the equilibrium point.

How does this change things? Does it change anything? Is it even correct?

Note: this is not intended to be a complete answer but is intended to assist the OP in understanding.


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