Two masses attached to a spring

In summary, the conversation discusses a problem involving two point masses attached to the ends of a spring with a spring constant of 'K'. The question is to find the maximum distance traveled by each block when the spring is released. The conversation also discusses whether the center of mass of the system will be at rest due to the absence of external forces. It is determined that the center of mass will not move due to the zero acceleration, and the maximum displacement of each block will be X0. The displacement of each block depends on the ratio of their masses, as stated in Newton's first law of motion.
  • #1
guru1323
2
0
I am stuck with a problem. There are two point masses with values m1 and m2 respectively. Both are attached to the two ends of a spring with spring constant 'K'. Initially the spring has a elongation = X0, then it is released to follow its natural motion. We have to find the maximum distance traveled by each of the two blocks. Since there are no external forces on the system, will the centre of mass of the system be at rest?
 
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  • #2
listen carefully.
since the two ends of the spring are connected to masses m1 and m2, hence you can consider the two halves of the string on either side of the centre of the string as two different springs connected to a fixed support (the centre).
each of the springs will have spring constant 2K(this can be obtained from some eqns.)
so you are right to think that the centre of mass of the system will be at rest.
however the is smt. wrong with the part of the question which says that the spring has an initial elongation X0.if it's so then X0 is the maxm. displacements of the block`
.
 
  • #3
guru1323 said:
Since there are no external forces on the system, will the centre of mass of the system be at rest?
Yes, that's the key. Hint: What will be the total displacement when the spring is compressed?
 
  • #4
anigeo said:
listen carefully.
since the two ends of the spring are connected to masses m1 and m2, hence you can consider the two halves of the string on either side of the centre of the string as two different springs connected to a fixed support (the centre).
each of the springs will have spring constant 2K(this can be obtained from some eqns.)
so you are right to think that the centre of mass of the system will be at rest.
however the is smt. wrong with the part of the question which says that the spring has an initial elongation X0.if it's so then X0 is the maxm. displacements of the block`
.

Doc Al said:
Yes, that's the key. Hint: What will be the total displacement when the spring is compressed?

Thank you both for ur replies...But i don't really understand the reason behind conclusion that COM will not move..I understand that acceleration of COM will be zero since there are no external forces present...but why can't we have a situation where COM moves with a constant velocity (acceleration = 0) till it comes to a maximum compression
@Anigo...X0 is the initial elongation of the whole spring and we have to calculate maximum displacements of the both blocks individually...I think sum of displacements of both blocks will be 2X0(Please correct me if i am wrong)
 
  • #5
guru1323 said:
But i don't really understand the reason behind conclusion that COM will not move..I understand that acceleration of COM will be zero since there are no external forces present...but why can't we have a situation where COM moves with a constant velocity (acceleration = 0) till it comes to a maximum compression
Since there are no external forces acting, the acceleration of the COM must be zero. Assuming it was released from rest, the COM cannot move.
I think sum of displacements of both blocks will be 2X0(Please correct me if i am wrong)
Sounds good to me.
 
  • #6
you r right in thinking that the COM might move with uniform velocity.but is there any difference between the state of rest and uniform motion.a uniform velocity does not alter or hamper our thing even we assume the COM to0 be at rest.
try to recall the Newtons's first law of motion.it says that there is no difference between the state of rest and uniform motion as the accn. in both the cases is 0.
moreover the maxm. displacement of each of the block will be X0.
 
  • #7
anigeo said:
moreover the maxm. displacement of each of the block will be X0.
The displacement of each block depends on the ratio of their masses.
 
  • #8
Doc Al said:
The displacement of each block depends on the ratio of their masses.

could u please explain me how does it happen?please procure the details.
 

FAQ: Two masses attached to a spring

What is the concept of "Two masses attached to a spring"?

The concept of "Two masses attached to a spring" is a common scenario in physics where two objects are connected by a spring and can move back and forth due to the force of the spring.

What is Hooke's Law and how does it relate to "Two masses attached to a spring"?

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. In the scenario of "Two masses attached to a spring", Hooke's Law explains the relationship between the force applied by the spring and the resulting displacement of the masses.

What factors affect the motion of "Two masses attached to a spring"?

The motion of "Two masses attached to a spring" is affected by several factors, including the masses of the objects, the spring constant, and the amplitude of the oscillations (i.e. how far the objects are displaced from their equilibrium position).

How is the period of oscillation calculated for "Two masses attached to a spring"?

The period of oscillation for "Two masses attached to a spring" can be calculated using the equation T = 2π√(m/k), where T is the period, m is the combined mass of the objects, and k is the spring constant.

Can "Two masses attached to a spring" be used to model real-world systems?

Yes, "Two masses attached to a spring" can be used to model various real-world systems, such as pendulums, vibrating systems, and shock absorbers. This simple system can provide insights into more complex systems and their behavior.

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