# Simple harmonic question

1. Feb 5, 2016

### erisedk

1. The problem statement, all variables and given/known data
Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad/s, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take g=10ms−2)

2. Relevant equations

3. The attempt at a solution
ω = √k/m
So, k = mω2 = 1×252 = 625
The maximum force will be exerted on the floor when the 1kg mass is at the lowest position, as the spring will be highly compressed and will push the bottom block.
So, total force on ground = 4g + kA + 1g
= 40 + 625*1.6*10-2 + 10
= 60 N

Now, someone asked me why I was adding the weight of the upper block ie. 1g as it isn't directly interacting with the 4 kg block. Only the spring force (and weight of 4kg block) are acting on the 4kg block, so the normal reaction should only depend on the two but not on the upper block. I don't know how to justify this even though I intuitively feel that the weight of the 1kg block should be added as it is part of the system. Can somebody provide a more rigorous explanation?

2. Feb 5, 2016

### haruspex

It is a common mistake to include forces as acting on a body which is not directly subject to the force. It is certainly the case that all you should count here is the weight of the lower block and the spring compression. But, have you correctly calculated the spring compression force? You multiplied k by the amplitude.
Suppose the upper block is only executing very small gentle oscillations. Will the spring ever be under tension? What will your kA method calculate as the compression in the spring when the upper block is at its highest point?

3. Feb 5, 2016

### erisedk

Thank you!! Got it.
The equilibrium position is itself shifted. Hence, when the system is in equilibrium, assuming that the spring is compressed by x0
kx0 = mg (m=1kg)
When it goes further down, i.e to the lowest position, the spring force on the block is
F = k(A+x0)
= kA + kx0
= kA + mg
This is why I was getting the right answer even though I hadn't really thought about it well.

4. Feb 5, 2016

Well done.