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

  1. Oct 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A moving particle displays simple harmonic motion with an amplitude of 3 cm. What position will it be for it to have half of its maximum speed?


    2. Relevant equations

    vmax= omega*A

    3. The attempt at a solution

    I divided the amplitude by 2 but I believe that's an incorrect method. was i suppose to deriviate x(t)=omega*A*sin(omega*t+phi)
     
  2. jcsd
  3. Oct 23, 2009 #2
    I'm going to assume that the particle acts like a mass on a spring. Then, we can solve this problem using conservation of energy.

    At the system's biggest displacement (3cm), there is no kinetic energy, so the total energy of the system is just the energy from the spring:

    [tex]E_{tot} = \frac{1}{2} k A^2[/tex]​

    At the equilibrium point, there is no spring energy. All of the energy is kinetic. Then, we can find the maximum velocity:

    [tex]E_{tot} = \frac{1}{2} m v_{max}^2[/tex]

    [tex]\frac{1}{2} k A^2 = \frac{1}{2} m v_{max}^2[/tex]

    [tex]\sqrt{\frac{k}{m}} A = v_{max}[/tex]​

    Note that [tex]\sqrt{k/m}[/tex] is usually called [tex]\omega[/tex], so this is simply the equation you gave us in your post.

    Now, we want to find the position of the mass when the velocity is half max. At that point, there is both spring energy and kinetic energy.

    [tex]E_{tot} = \frac{1}{2} k x^2 + \frac{1}{2} m \left( \frac{v_{max}}{2} \right)^2[/tex]

    [tex]\frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2 \cdot 4} m \frac{k}{m} A^2 [/tex]

    [tex]A^2 = x^2 + \frac{1}{4} A^2 [/tex]

    [tex]\frac{3}{4} A^2 = x^2[/tex]

    [tex]\sqrt{\frac{3}{4}} A = x[/tex]​

    So, the position of the mass when it's traveling at half the maximum speed is [tex]x = \sqrt{\frac{3}{4}} A[/tex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple harmonic motion
Loading...