1. Not finding help here? Sign up for a free 30min 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!

Simple Harmonic Motion of a block

  1. Mar 29, 2006 #1
    A 0.50 kg block sits on top of a 5.00-kg block that rests on the floor. The larger block is attached to a horizontal spring that has a force constant of 15.0 N/m. It is displaced and undergoes simple harmonic motion. What is the largest possible amplitude the 5.00-kg mass can have for the smaller mass to remain at rest relative to the larger block? The coefficient of static friction between the two blocks is 0.25. There is no friction between the larger block and the floor.

    I have all the equations:

    [tex] x(t) = A\cos(\omega(t) + \phi) [/tex]
    [tex] v(t) = -A\omega\sin(\omega(t)+\phi) [/tex]
    [tex] a(t) = -A\omega^{2}\cos(\omega(t)+\phi) [/tex]
    [tex] T = \frac{1}{f} [/tex]
    [tex] \omega = \sqrt{\frac{k}{m}} [/tex]
    [tex] T = 2\pi\sqrt{\frac{m}{k}} [/tex]

    I drew a free body diagram. I know that the velocity of the smaller block has to be 0. The force the spring exerts on the lower block has to be less than the friction force for the velocity to be 0.

    Any help would be appreciated.

    note: It should be [tex] \omega(t) [/tex]

    Thanks
     
    Last edited: Mar 29, 2006
  2. jcsd
  3. Mar 29, 2006 #2
    Do we not also need the dimensions of the contact surface between the two blocks? The static friction force is dependant on this information isnt it?
     
  4. Mar 29, 2006 #3
    I determined the static friction force to be [tex] f_{s} = (0.25)(0.5 kg)(9.8\frac{m}{s^{2}}) = 1.225 N [/tex]

    So do I just set this equal to:
    [tex] 1.225 N = (15.0 \frac{N}{m} )(x) [/tex], and solve for x?

    thanks
     
  5. Mar 29, 2006 #4

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No! Why would that be true??
    What you must do is to focus on the smaller block. You must have [itex] \sum F_x = m a_x [/itex]. The only force in the x direction is the friction force (the normal force cancels gravity). Now, you know that the friction force must be smaller or equal to [itex] \mu_s N [/itex]. Therefore, you can find the maximum acceleration....Once you know the maximum acceleration of the two blocks, you are almost there...
     
  6. Mar 30, 2006 #5

    Astronuc

    User Avatar

    Staff: Mentor

    One has to determine the maximum lateral acceleration that the blocks can experience without the smaller block moving.

    You found the friction force, Fs between the smaller block and the larger block. Now, the lateral for F = ma = Fs for the small block to remain stationary. With ma = [itex]\mu\,mg[/itex], one finds that

    a = [itex]\mu\,g[/itex].

    From the spring equation F = M a = k x, one can find that the maximum force (and therefore maximum acceleration) corresponds to the maximum displacement (amplitude).

    Thus x = Ma/k, but the key here is on what mass M is the spring acting.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?