A few questions on different spring and mass setups

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    Mass Spring
In summary: The spring constant can be derived from the frequency. In summary, the force exerted by two springs on a mass attached to them is half of the force exerted by a single spring displaced by the same amount. For a system with two masses attached to a spring, the force exerted on each mass is the same as that of an unmoving wall. To find the spring constant of a loaded spring, the force exerted by the weight is divided by the displacement caused by the weight. However, in practice, this method may be affected by the oscillation of the system. Another method of finding the spring constant is by measuring the frequency of the oscillation.
  • #1
chipotleaway
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One mass, two springs

If you have a mass attached to a spring which is attached to another spring which both have the same spring constant, why is the force exerted by the two springs on the mass [itex]F=-\frac{1}{2}kx[/itex]?

Is it because when you compare it to a single spring displaced by the same Δx, each spring in the two spring system has only been stretched by half, and they act like a single spring? (not too clear on my reasoning)Two masses, one spring
If two masses are attached on either end of a spring and it's stretched, would the force on each mass be half of what it would if the other end were attached to a wall?

Loaded spring
There's an example in my textbook that presents the following situation: you have a traincar of 1000kg on a spring, it's loaded slowly with a weight of 980N which causes the spring to be compressed by some 0.28m and the system then undergoes simple harmonic motion.

To find the spring constant, they divided the force 980N by the displacement, but doesn't Hooke's law relate the force from the spring on whatever's pulling/pushing it? If the spring pushes back with 980N when displaced by 0.28m, then shouldn't the thing be in equilibrium and not undergo SHO at all?

Thanks
 
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  • #2
chipotleaway said:
One mass, two springs

If you have a mass attached to a spring which is attached to another spring which both have the same spring constant, why is the force exerted by the two springs on the mass [itex]F=-\frac{1}{2}kx[/itex]?

Is it because when you compare it to a single spring displaced by the same Δx, each spring in the two spring system has only been stretched by half, and they act like a single spring? (not too clear on my reasoning)
Yes, that's pretty much right. It may be easier to imagine that the two springs are connected to either end of a massless block instead of directly connected; then you can continue thinking about them as two separate springs each with their own stretched length.

Two masses, one spring
If two masses are attached on either end of a spring and it's stretched, would the force on each mass be half of what it would if the other end were attached to a wall?
No. As long as the two masses aren't moving the force they're exerting on the spring, and therefore the force the spring is exerting on them, is the same as that of an unmoving wall. (This is also the answer if one or both masses are allowed to move - the only difference is that the stretch of the spring and hence the forces will change over time as the masses move)
Loaded spring
There's an example in my textbook that presents the following situation: you have a traincar of 1000kg on a spring, it's loaded slowly with a weight of 980N which causes the spring to be compressed by some 0.28m and the system then undergoes simple harmonic motion.

To find the spring constant, they divided the force 980N by the displacement, but doesn't Hooke's law relate the force from the spring on whatever's pulling/pushing it? If the spring pushes back with 980N when displaced by 0.28m, then shouldn't the thing be in equilibrium and not undergo SHO at all?

If you could push it down exactly .28 meters and then hold it exactly perfectly still while you let go of it then it would be in equilibrium and not oscillate, just as you say. In practice no one has hands that steady - you'll let go of it a bit too high or a bit too low, or give it just a teeny extra push, and that will set it to harmonic oscillation.
 
  • #3
Thanks. For the third one, I really wanted to know how valid it was to find the spring constant that way
 
  • #4
chipotleaway said:
Thanks. For the third one, I really wanted to know how valid it was to find the spring constant that way

It's a valid method, as long as you're careful not to let the inevitable oscillation throw off your distance measurement. Waiting for the oscillation to die down might be the easiest way of doing this.

As an aside, another way of finding a spring constant is to deliberately start the thing oscillating, then measure the frequency of the oscillation.
 
  • #5
for your questions about different spring and mass setups. Let me address each one individually:

1. One mass, two springs: The force exerted by the two springs on the mass is equal to the sum of the individual forces exerted by each spring. In this case, the force exerted by each spring is half of the total force because they both have the same spring constant and are stretched by the same amount. This is because the springs are acting in parallel, meaning they are both connected to the same mass. The equation F=-\frac{1}{2}kx is a result of this parallel arrangement and the fact that the two springs act together as a single spring with half the spring constant.

2. Two masses, one spring: In this setup, the force on each mass would not necessarily be half of what it would be if the other end were attached to a wall. This is because the spring is not acting in parallel, but in series. This means that the force exerted by the spring is divided between the two masses, depending on their individual masses and displacements. The equation F=-kx can still be used to calculate the force on each mass, but it may not be exactly half of what it would be with one mass attached.

3. Loaded spring: The textbook example you mentioned is using the concept of Hooke's law to find the spring constant of the spring. The force exerted by the weight is equal to the force exerted by the spring, as the system is in equilibrium. 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 this case, the spring is compressed by 0.28m, so the force exerted by the spring is equal to 980N. Dividing this force by the displacement gives us the spring constant, which is 3500 N/m in this case. This is the same concept used in the previous examples, but in this case, the system is not undergoing simple harmonic motion. The spring is in equilibrium, meaning the forces are balanced and there is no SHO.
 

What is the purpose of studying different spring and mass setups?

The purpose of studying different spring and mass setups is to understand the behavior and dynamics of these systems, which can be applied in various fields such as engineering, physics, and mechanics.

What are the factors that affect the behavior of a spring and mass system?

The factors that affect the behavior of a spring and mass system include the stiffness of the spring, the mass of the object attached to the spring, and the amplitude and frequency of the oscillations.

How does the stiffness of a spring affect the behavior of a spring and mass system?

The stiffness of a spring determines the amount of force required to stretch or compress the spring, which in turn affects the amplitude and frequency of the oscillations of the system.

What is the difference between a simple harmonic motion and a damped harmonic motion?

In a simple harmonic motion, the amplitude and frequency of the oscillations remain constant, while in a damped harmonic motion, the amplitude decreases over time due to the presence of a damping force.

How can the behavior of a spring and mass system be manipulated?

The behavior of a spring and mass system can be manipulated by changing the stiffness of the spring, altering the mass of the object, or adjusting the amplitude and frequency of the oscillations.

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