1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A few questions on different spring and mass setups

  1. Oct 23, 2013 #1
    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
     
    Last edited: Oct 23, 2013
  2. jcsd
  3. Oct 23, 2013 #2

    Nugatory

    User Avatar

    Staff: Mentor

    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.

    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)
    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.
     
  4. Oct 25, 2013 #3
    Thanks. For the third one, I really wanted to know how valid it was to find the spring constant that way
     
  5. Oct 25, 2013 #4

    Nugatory

    User Avatar

    Staff: Mentor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook