# New amplitude of a spring-mass system when a 2nd mass falls on the first

• songoku
In summary, the conversation discusses the confusion about the conservation of momentum in a system with two objects, one performing simple harmonic motion and the other falling vertically. It is concluded that the momentum is not conserved in either the vertical or horizontal direction due to external forces acting on the system. However, if the impulse of the external force is small enough, conservation of momentum in the horizontal direction can be approximated and used to solve the problem. This is due to the assumption that the collision between the two objects is completed in a very short time interval where the external force has no effect.
songoku
Homework Statement
A mass m is connected to a spring of spring constant k and performs simple harmonic motion with amplitude A on a horizontal surface. When the spring is stretched by A/2, another mass m falls vertically on first mass. What is the new amplitude of motion?
Relevant Equations
Simple harmonic motion

Conservation of momentum?
I am confused about conservation of momentum for this question.

I take the system as two objects, each having mass m. For first object (the one performing simple harmonic motion), there is net external force acting on it in horizontal direction (restoring force of spring). For second object (the one falling vertically), there is net external force acting on it (its weight).

1. The momentum in vertical direction is not conserved because initially (before they stuck), there is initial vertical momentum but after they stuck together, the vertical momentum is zero. But how about momentum in horizontal direction. I think momentum in horizontal direction is also not conserved because there is net external force in that direction. Or we can think it is also conserved because before and after they stuck there are net external force (and assuming these forces have equal magnitude)?

2. Do we need to consider momentum to solve this question?

Thanks

Delta2
Yes, you need to conserve horizontal momentum otherwise you will not be able to solve the problem. You may assume that the collision is completed in time ##\Delta t## much smaller than the period of oscillations so the horizontal momentum of the system before and after the collision is the same and so is the external force of the spring. In other words, assume that the oscillating mass magically doubles suddenly when its displacement is ##A/2##.

Delta2 and Lnewqban
kuruman said:
Yes, you need to conserve horizontal momentum otherwise you will not be able to solve the problem. You may assume that the collision is completed in time ##\Delta t## much smaller than the period of oscillations so the horizontal momentum of the system before and after the collision is the same and so is the external force of the spring. In other words, assume that the oscillating mass magically doubles suddenly when its displacement is ##A/2##.

What I know about law of conservation of momentum is "the total momentum of a system is constant if no net external force acting on the system".

So the law should be: ""the total momentum of a system is constant if there is no change in net external force acting on the system"? Or maybe I miss some basic concept of momentum?

Thanks

songoku said:
What I know about law of conservation of momentum is "the total momentum of a system is constant if no net external force acting on the system".

So the law should be: ""the total momentum of a system is constant if there is no change in net external force acting on the system"? Or maybe I miss some basic concept of momentum?

Thanks
No the law of conservation of momentum is fine as it is.

For this problem the momentum is not conserved neither in the vertical or the horizontal direction:

• in the vertical direction you have the reaction from the "floor" that increases during the collision time ##\Delta t## and its impulse "neutralizes" the (vertical momentum of the falling mass + the impulse of the weight of the falling mass)
• in the horizontal direction you have the external force from the spring that will impart horizontal impulse ##\Omega=\int_{t_0}^{t_0+\Delta t} kx(t)dt## during the collision time ##\Delta t## and this will make the horizontal momentum to not be conserved
So you will not be able to solve this problem using conservation of momentum. However if we assume that the impulse ##\Omega## is small enough (in comparison with the momentum that the oscillating mass has at the time instant ##t_0## that the collision starts and in comparison with the impulse of the force that arise during the collision) then conservation of momentum in the horizontal direction holds approximately and so you can use it to solve the problem .

Last edited:
etotheipi and Lnewqban
Delta2 said:
No the law of conservation of momentum is fine as it is.

For this problem the momentum is not conserved neither in the vertical or the horizontal direction:

• in the vertical direction you have the reaction from the "floor" that increases during the collision time ##\Delta t## and its impulse "neutralizes" the (vertical momentum of the falling mass + the impulse of the weight of the falling mass)
• in the horizontal direction you have the external force from the spring that will impart horizontal impulse ##\Omega=\int_{t_0}^{t_0+\Delta t} kx(t)dt## during the collision time ##\Delta t## and this will make the horizontal momentum to not be conserved
So you will not be able to solve this problem using conservation of momentum. However if we assume that the impulse ##\Omega## is small enough (in comparison with the momentum that the oscillating mass has at the time instant ##t_0## that the collision starts) then conservation of momentum in the horizontal direction holds approximately and so you can use it to solve the problem .

I see.Thank you very much kuruman and delta2

kuruman and Delta2
Just to summarize what @Delta2 said in post #4, the second mass has zero initial horizontal velocity when the first mass is moving horizontally while experiencing an external force of magnitude ##kA/2##. The approximation is that when the first and second masses reach their new common horizontal velocity, the external force on the first mass is still ##kA/2##. There is a subtle difference between what this implies and what you inferred from it,
songoku said:
So the law should be: ""the total momentum of a system is constant if there is no change in net external force acting on the system"? Or maybe I miss some basic concept of momentum?
When there is an external force acting on the system, momentum is not conserved, period. What we are asserting here is that the collision is completed so fast that the external force, although present, has no effect on the system while the collision is taking place. When a force has no effect on a system, it's as if the force does not exist. Therefore, fully aware that this is only an approximation, we suspend the external force during the very short collision-time interval which then allows us to conserve momentum and solve the problem.

songoku, etotheipi, Lnewqban and 1 other person
Deleted.

Last edited:
Lnewqban said:
The velocity of the previously oscillating mass was zero at point A/2, exactly when the addition of masses happened, according to the original text of the problem.
I am not sure why you are saying this. If the second mass didn't drop on it, it would keep on going until reaching full amplitude at ##x=A##. The original text of the problem says that the amplitude is ##A##.

Lnewqban
kuruman said:
I am not sure why you are saying this. If the second mass didn't drop on it, it would keep on going until reaching full amplitude at ##x=A##. The original text of the problem says that the amplitude is ##A##.
That is correct.
Hoping that the problem was simplified that way, I misread the part stating "When the spring is stretched by A/2".
Editing previous post.
Apologies.

kuruman

## 1. What is the formula for calculating the new amplitude of a spring-mass system when a second mass falls on the first?

The formula for calculating the new amplitude of a spring-mass system when a second mass falls on the first is A = (m1 + m2)g/k, where m1 and m2 are the masses of the first and second objects, g is the acceleration due to gravity, and k is the spring constant.

## 2. How does the addition of a second mass affect the amplitude of a spring-mass system?

The addition of a second mass to a spring-mass system will increase the amplitude of the system. This is because the second mass adds to the overall weight of the system, causing the spring to stretch further and thus increasing the amplitude.

## 3. Does the mass of the second object have a significant impact on the new amplitude of the system?

Yes, the mass of the second object does have a significant impact on the new amplitude of the system. The larger the mass, the greater the increase in amplitude will be. This is because a larger mass will add more weight to the system, causing the spring to stretch further.

## 4. Is the new amplitude of the system affected by the spring constant?

Yes, the new amplitude of the system is affected by the spring constant. A higher spring constant will result in a smaller amplitude increase, while a lower spring constant will result in a larger amplitude increase. This is because a higher spring constant means the spring is stiffer and will not stretch as much when additional weight is added.

## 5. Can the new amplitude of the system be calculated if the mass and spring constant are known?

Yes, the new amplitude of the system can be calculated if the mass and spring constant are known. The formula A = (m1 + m2)g/k can be used to calculate the new amplitude, as long as the values for m1, m2, g, and k are known.

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