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.

| cite | improve this answer | |

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.