Results 1 to 2 of 2

Thread: Harmonic Motion

  1. #1
    Newbie
    Joined
    Aug 2019
    From
    Dhaka
    Posts
    1

    Cool Harmonic Motion

    How to solve this:
    6 a Calculate the displacement in terms of x0, expressed as a fraction of the amplitude x0, of a particle moving in simple harmonic motion with a speed equal to half the maximum value.
    b Calculate the displacement at which the energy of the particle has equal amounts of kinetic and of potential energy
    I am stuck at both (a) and (b) and just don't know how to do this. Thanks for your help in advance!.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Cervesa's Avatar
    Joined
    Dec 2014
    From
    USA
    Posts
    94
    Thanks
    74

    Re: Simple Harmonic Motion

    Quote Originally Posted by Mark739 View Post
    How to solve this:
    6 a Calculate the displacement in terms of x0, expressed as a fraction of the amplitude x0, of a particle moving in simple harmonic motion with a speed equal to half the maximum value.
    b Calculate the displacement at which the energy of the particle has equal amounts of kinetic and of potential energy
    I am stuck at both (a) and (b) and just don't know how to do this. Thanks for your help in advance!.
    assuming the object undergoing SHM starts at maximum displacement, $x_0$

    $x(t) = x_0 \cos(\omega t)$, where $\omega = \dfrac{2\pi}{T}$

    (a) $x'(t) = v(t) = -x_0 \cdot \omega \sin(\omega t) \implies |v_{max}| = x_0 \cdot \omega$

    at half max speed, $\dfrac{x_0 \cdot \omega}{2} = |v(t)| = x_0 \cdot \omega \sin(\omega t) \implies \sin(\omega t) = \dfrac{1}{2} \implies \omega t = \dfrac{\pi}{6} \implies t = \dfrac{\pi}{6\omega}$

    $x\left(\dfrac{\pi}{6\omega}\right) = x_0 \cos \left(\omega \cdot \dfrac{\pi}{6\omega}\right) = \dfrac{x_0 \sqrt{3}}{2}$


    (b) Max kinetic energy (equal to maximum mechanical energy of the system) occurs at equilibrium and is $K_{max} = \dfrac{1}{2}m v_{max}^2 = \dfrac{1}{2}m (x_0 \omega)^2$

    When kinetic and potential energy are equal, $K = \dfrac{1}{2}m|v(t)|^2 = \dfrac{1}{4}m (x_0 \omega)^2 \implies x_0 \omega \cos(\omega t) = \dfrac{x_0 \omega}{\sqrt{2}} \implies \omega t = \dfrac{\pi}{4} \implies t = \dfrac{\pi}{4\omega}$

    $x\left(\dfrac{\pi}{4\omega} \right) = x_0 \cos\left(\omega \cdot \dfrac{\pi}{4\omega} \right) = \dfrac{x_0}{\sqrt{2}}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. harmonic motion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 7th 2014, 09:52 PM
  2. Harmonic motion
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 22nd 2013, 06:09 AM
  3. Harmonic Motion
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: Aug 6th 2012, 02:55 PM
  4. harmonic motion??
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Feb 4th 2009, 05:27 PM
  5. Harmonic Motion
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jun 30th 2008, 03:39 PM

/mathhelpforum @mathhelpforum