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!

Simple harmonic motion platform

  1. Sep 11, 2014 #1
    1. A platform is executing simple harmonic motion in a vertical direction with an amplitude of 5 cm and a frequency of 10/pi vibrations per second. a block is placed on the platform at the lowest point of its path.
    a) at what point will the block leave the platform?
    b)how far will the block reach above the highest point that the platform reaches

    2. Relevant equations

    x=Acos(ωt-phi)
    mg=-ma(where the block leaves the platform)
    g-(ω^2)Acos(ωt)=0
    ω=2pif
    m(g-(ω^2)Acos(ωt))=0

    3. The attempt at a solution for part a

    given the frequency i found the angular frequency to be 20rad/s

    knowing x=-Acos(ωt) because it starts out at the lowest point at -A and also no phase angle to the equation.

    to find when the mass leaves the platform -ma=mg, and taking the derivative twice of this equation a=(ω^2)Acos(ωt)

    and i need to satisfy the initial condition of the acceleration so i get m(g-(ω^2)Acos(ωt))=0

    but how do find when this happens?

    part b I was not able to get to yet
     
    Last edited: Sep 11, 2014
  2. jcsd
  3. Sep 11, 2014 #2

    nrqed

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Don't they ask the position (they say "at what point"), not the time? Then you just need - Acos(ωt), no?
     
  4. Sep 11, 2014 #3
    Just to let people know I actually found out how to do the problem sorry for posting this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Simple harmonic motion platform
Loading...