A few questions on different spring and mass setups

[chipotleaway]

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 F=-\frac{1}{2}kx?

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

 

[Nugatory]

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 F=-\frac{1}{2}kx?

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.

 

[chipotleaway]

Thanks. For the third one, I really wanted to know how valid it was to find the spring constant that way

 

[Nugatory]

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.