# A mass is dropped on a plate that's suspended from a spring

## Homework Statement

a load of mass m falls a height h onto a pan hung from a spring. if the spring constant is k and the pan is massless and m does not bounce, the amplitude of oscillation is

F = - kx

U = 1/2kx^2

## The Attempt at a Solution

mgh = 1/2kx^2, x = sqrt(2mgh/k).this is total extension. Now mg/k will give us the mean position

sqrt(2mgh/k) - mg/k should be the amplitude? Well
its wrong though :|

kuruman
Homework Helper
Gold Member
Consider mechanical energy conservation from point A (where the mass first makes contact) to point B where the mass is at the lowest point of its motion. Write ΔK + ΔU = 0. Note that ΔU has two terms, spring and gravity.

Consider mechanical energy conservation from point A (where the mass first makes contact) to point B where the mass is at the lowest point of its motion. Write ΔK + ΔU = 0. Note that ΔU has two terms, spring and gravity.
okay, I gotit thanks!

@kuruman , the setup of the problem is quite interesting , in the sense that even though collision between mass and pan is inelastic , the total mechanical energy is conserved .

I find this bit unusual .

kuruman
Homework Helper
Gold Member
@kuruman , the setup of the problem is quite interesting , in the sense that even though collision between mass and pan is inelastic , the total mechanical energy is conserved .
I find this bit unusual .
Yes, I think the author of the problem made the pan massless to simplify it. It's the same situation as the more commonly seen problem of a mass colliding with a spring - no platform. If the pan has mass, then momentum conservation gives the speed of the oscillator as ##v_A = (m~\sqrt{2gh})/(m+m_{pan})##. At this point one would have to solve
$$\frac{1}{2}k\left[ \frac{(m+m_{pan})g}{k} \right]^2+\frac{1}{2}(m+m_{pan})\left[ \frac{m~\sqrt{2gh}}{m+m_{pan}} \right]^2=\frac{1}{2}kA^2$$
for ##A##. In the limit ##m_{pan} \rightarrow 0,## one gets the same equation as would be obtained with energy conservation. I find that comforting.

conscience
Yes, I think the author of the problem made the pan massless to simplify it. It's the same situation as the more commonly seen problem of a mass colliding with a spring - no platform. If the pan has mass, then momentum conservation gives the speed of the oscillator as ##v_A = (m~\sqrt{2gh})/(m+m_{pan})##. At this point one would have to solve
$$\frac{1}{2}k\left[ \frac{(m+m_{pan})g}{k} \right]^2+\frac{1}{2}(m+m_{pan})\left[ \frac{m~\sqrt{2gh}}{m+m_{pan}} \right]^2=\frac{1}{2}kA^2$$
for ##A##. In the limit ##m_{pan} \rightarrow 0,## one gets the same equation as would be obtained with energy conservation. I find that comforting.

Nice analysis !

jbriggs444
Homework Helper
m does not bounce
How are you reading this clause? That m does not bounce from the pan on impact? Or that m does not leave the pan after the oscillation has returned it to the impact point (i.e. glue on the surface of the pan)?

kuruman