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Proving the various forms are equivalent

  1. Nov 18, 2013 #1
    1. The problem statement, all variables and given/known data
    In Section 5.2 we discussed four equivalent ways to represent simple harmonic motion in one
    dimension:

    [itex]x(t) = C_1 e^{i \omega t} + C_2 e^{-i \omega t}[/itex] (1)

    [itex]= B_1 cos(\omega t) + B_2 sin (\omega t)[/itex] (2)

    [itex]= A cos(\omega t - \delta)[/itex] (3)

    [itex]=Re C e^{i \omega t}[/itex] (4)

    To make sure you understand all of these, show that they are equivalent by proving the following implications: I → II → III → IV → I. For each form, give an expression for the constants (C1, C2, etc.) in terms of the constants of the previous form.


    I was successful in showing that (2) follows from (1). However, know I am endeavoring to show that (3) follows from (2).

    At first, I thought that there might be some trigonometric identity involving the sum of two trig functions; but I couldn't not find any such identity. Then I tried to write the sine function in terms of cosine, noting that the sine function is shifted 90 degrees; this, too, did not seem very helpful. Finally, I noted that both trig functions has the same argument, from which I tried to draw a triangle, but immediately realized that this would be of no help.

    Could anyone provide a hint?
     
  2. jcsd
  3. Nov 18, 2013 #2

    Curious3141

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    Homework Helper

    Expand (3) using the angle sum formula for cosine. Compare with (2). Set up a pair of simultaneous equations relating A and δ to B1 and B2.

    To solve them for A and δ, there is a well-known trick using trig identities. First square the two equations and add. Remember that ##\sin^2\delta + \cos^2\delta = 1##. That should get you A quickly.

    To solve for δ, divide the two equations, cancelling out A. Remember that ##\frac{\sin\delta}{\cos\delta} = \tan\delta##.
     
  4. Nov 19, 2013 #3
    I am not sure that is correct. I need to show that (3) results from (2) through manipulations and definitions, not the other way around.
     
  5. Nov 19, 2013 #4

    Curious3141

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    Currently A and δ are undefined in terms of B1 and B2. What you need to do is to show that you can solve for these variables in terms of the earlier ones. Then you *have* shown that (2) can be re-expressed as (3).
     
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