Conservation of momentum and SHM

In summary, the conversation was about two point masses coupled by a spring and their oscillatory motion after the spring breaks. The displacement of one mass was given and the task was to find the displacement of the other mass and the relationship between the amplitude of the oscillations and the uncompressed length of the spring. The solution involved conserving energy at two different times and setting the maximum gap between the two masses equal to the maximum stretching of the spring plus its natural length. The resulting equation shows that the amplitude is directly proportional to the sum of the masses and inversely proportional to the mass of the second mass.
  • #1
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Homework Statement


Two point masses m1 and m2 are coupled by a spring of spring constant k and uncompressed length L0. The spring is fully compressed and a thread ties the masses together with negligible separation between them. The tied assembly is moving in the +x direction with uniform speed v0. At a time, say t = 0, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring, start oscillating. The displacement of mass m1 is given by x1 (t) = v0t - A(1-cos(ωt)) where A is a constant. Find (i) the displacement x2(t) of mass m2, and (ii) the relationship between A and L0.

Homework Equations

The Attempt at a Solution


First part is easy. Using
##x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}##
and substituting ##x_{cm} = v_0t## and ##x_1= v_0t - A(1-\cos{ωt})##
we get ##x_2 = v_0t + \dfrac{m_1}{m_2}.A(1-\cos{wt})##

However, I'm not sure what to do for part (ii). I suppose it involves using the energy equation, but that isn't really working out because of the ##t## (time). I think we might have to minimise or maximise something, in any case, I'm not sure how to proceed. Please help.
 
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  • #2
erisedk said:

Homework Statement


Two point masses m1 and m2 are coupled by a spring of spring constant k and uncompressed length L0. The spring is fully compressed and a thread ties the masses together with negligible separation between them. The tied assembly is moving in the +x direction with uniform speed v0. At a time, say t = 0, it is passing the origin and at that instant the thread breaks. The masses, attached to the spring, start oscillating. The displacement of mass m1 is given by x1 (t) = v0t - A(1-cos(ωt)) where A is a constant. Find (i) the displacement x2(t) of mass m2, and (ii) the relationship between A and L0.

Homework Equations

The Attempt at a Solution


First part is easy. Using
##x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}##
and substituting ##x_{cm} = v_0t## and ##x_1= v_0t - A(1-\cos{ωt})##
we get ##x_2 = v_0t + \dfrac{m_1}{m_2}.A(1-\cos{wt})##

However, I'm not sure what to do for part (ii). I suppose it involves using the energy equation, but that isn't really working out because of the ##t## (time). I think we might have to minimise or maximise something, in any case, I'm not sure how to proceed. Please help.
Energy should do it. Please post your working.
 
  • #3
Sorry. I forgot to post the working.
Conserving energy at t=0 and t=t,

##\dfrac{1}{2}.k{L_0}^2 + \dfrac{1}{2}(m_1 + m_2){v_0}^2 = \dfrac{1}{2}k{(L_0 - |x_2 - x_1|)}^2 + \dfrac{1}{2}m_1{v_1}^2 + \dfrac{1}{2}m_2{v_2}^2##

where ##v_1 = v_0 - Aω\sin{ωt}## and ##v_2 = v_0 + A\dfrac{m_1}{m_2}ω\sin{ωt}##

This is just terrible.
So, I'm thinking there is definitely some optimisation involved, something like ##\sin{ωt} = 0## or ## \cos{ωt} = 0## but I'm not seeing what.
 
  • #4
erisedk said:
Sorry. I forgot to post the working.
Conserving energy at t=0 and t=t,

##\dfrac{1}{2}.k{L_0}^2 + \dfrac{1}{2}(m_1 + m_2){v_0}^2 = \dfrac{1}{2}k{(L_0 - |x_2 - x_1|)}^2 + \dfrac{1}{2}m_1{v_1}^2 + \dfrac{1}{2}m_2{v_2}^2##

where ##v_1 = v_0 - Aω\sin{ωt}## and ##v_2 = v_0 + A\dfrac{m_1}{m_2}ω\sin{ωt}##

This is just terrible.
So, I'm thinking there is definitely some optimisation involved, something like ##\sin{ωt} = 0## or ## \cos{ωt} = 0## but I'm not seeing what.
It is not a question of optimisation. You need to pick a particular time t to compare with t=0.
 
  • #5
(ii) maximum compression = maximum stretching = L
So max. gap btw. x1 and x2 = max. Stretch + natural length = L + L = 2L
x2 - x1 = Am1/m2(1-coswt) + A(1-coswt)
(x2 - x1) max. =» 2A(m1/m2) + 2A = 2L
L = A(m1 + m2)/m2
 
  • #6
Shivam aditya said:
(ii) maximum compression = maximum stretching = L
So max. gap btw. x1 and x2 = max. Stretch + natural length = L + L = 2L
x2 - x1 = Am1/m2(1-coswt) + A(1-coswt)
(x2 - x1) max. =» 2A(m1/m2) + 2A = 2L
L = A(m1 + m2)/m2

The original post was from 3 years ago!
 

What is the law of conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum of all objects before a collision or interaction is equal to the total momentum after the collision or interaction. This means that momentum is conserved, or remains constant, in a closed system.

How does conservation of momentum apply to simple harmonic motion (SHM)?

In SHM, the restoring force is directly proportional to the displacement from equilibrium. This means that the force acting on the object is always changing, but the total momentum of the system remains constant. As a result, the velocity and displacement of the object also vary sinusoidally.

What is an example of conservation of momentum in SHM?

One example is a simple pendulum. As the pendulum swings back and forth, the total momentum of the pendulum and the Earth remains constant. The object gains momentum as it moves towards the bottom of its swing, and loses momentum as it moves towards the top, but the total momentum remains the same.

How does conservation of momentum affect collisions?

In collisions, the total momentum of the objects involved remains constant. This means that the sum of the momenta of the objects before the collision is equal to the sum of the momenta after the collision. This can be seen in both elastic and inelastic collisions.

Can conservation of momentum be violated?

No, the law of conservation of momentum is a fundamental law of physics and has been proven to hold true in countless experiments. However, it can seem like momentum is not conserved in certain situations, but this is typically due to external forces that are not taken into account or are not part of the closed system.

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