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Homework Help: Simple harmonic motion (Again)

  1. Feb 5, 2014 #1
    Simple harmonic motion (Again) :(

    This is not a question about a problem it is more about the position of a simple harmonic oscillator as a function of time:)

    I went through it in a lecture yesterday and found using the energy in simple harmonic motion to be

    x(t)= A cos(ωt +φ)

    Which is fine the proof seems ok but when I look through my university text book for some supplementary work on the subject I found that the book had something slightly different.

    x(t)= A sin(ωt +φ)

    Why is this that case?

    I should also say I googled and looked in other books and found a mixture of the two equations although I did notice that a lot of the time when it was written sine, the argument didn't contain a phase constant.....

    x(t)= A sin(ωt)

    Which makes me think it has something to do with the phase constant but I'm not sure because my university text book includes the phase constant and also the sine of the argument it's contained in:)

    Thanks for your help in advanced
  2. jcsd
  3. Feb 5, 2014 #2
    Brush up on your trigonometry. $$\overbrace {\pm}^{?} \cos \left(\overbrace {\pm}^{?}\alpha + ? \right) = \overbrace {\pm}^{?} \sin \left(\overbrace {\pm}^{?} \alpha + ? \right) $$
  4. Feb 5, 2014 #3
    Hello, KiNGGeexD here's the solution for your probem.

    The Sin and the Cosine functions depends upon the position of the object with respect to the observer. When you look at the derivation of the displacement equation, you get the equation to be
    x = A sin (wt) or more specifically x = A sin (wt + Z) where Z is your phase constant. Now, let us see why that phase constant comes into picture.

    Suppose two pendula are performing simple harmonic motion about their mean position. Now supose, the second pendulum starts oscillating when the first pendulum is at its maximum displacement. So, naturally second pendulum lags behind in phase with respect to first pendulum. So, when we write the equation of the second pendulum, we include a phase constant in the equation which implies that the particle exceeds or lags in phase. So, your equation comes out to be
    x = A Sin(wt + Z)

    About the two equations.

    when x = A sin (wt) it implies that the particle is starting its motion from the mean position. ......(1)

    when x = A cos (wt) it implies that the particle is starting its motion from extreme positions. It can be viewed as if the particle has gone to extreme position from mean position.(In this case its opposite, the particle starts from extreme position) when particle goes to extreme position it has completed half cycle which corresponds to ∏/2 radiance. if you substitute in equation (1) you will get your equation as x = A cos (wt). So, it doesn't matter whichever equation you use unless and until you are working in correct regime.

    Devendra S. Chavan
  5. Feb 5, 2014 #4
    That makes sense! Thanks!
    So I would of course have different derivatives dependent on the situation?

  6. Feb 5, 2014 #5
    Would be correct to say that using cosine or sine is just like setting a coordinate system? It is merely a reference to the initial conditions of a particular scenario?

    So it really doesn't make a difference?
  7. Feb 5, 2014 #6
    If sin is being used we must add pi/2 to the phase constant?
  8. Feb 6, 2014 #7
    See, the phase constant can be anything, any angle. So, it is not necessary to add pi/2 always. If you subtract pi/2 in the above equation, you still get the same answer but with opposite sign indicating that the motion is along the -ve side of the cartesian co-ordinate system.
  9. Feb 6, 2014 #8
    Ok thanks that helps clear it up a little! I was thinking about it in an incorrect way
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