Proving the various forms are equivalent

[embphysics]

Homework Statement

In Section 5.2 we discussed four equivalent ways to represent simple harmonic motion in one
dimension:

$x(t) = C_1 e^{i \omega t} + C_2 e^{-i \omega t}$ (1)

$= B_1 cos(\omega t) + B_2 sin (\omega t)$ (2)

$= A cos(\omega t - \delta)$ (3)

$=Re C e^{i \omega t}$ (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?

 

[Curious3141]

Homework Statement

In Section 5.2 we discussed four equivalent ways to represent simple harmonic motion in one
dimension:

$x(t) = C_1 e^{i \omega t} + C_2 e^{-i \omega t}$ (1)

$= B_1 cos(\omega t) + B_2 sin (\omega t)$ (2)

$= A cos(\omega t - \delta)$ (3)

$=Re C e^{i \omega t}$ (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?

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 $\frac{\sin\delta}{\cos\delta} = \tan\delta$.

 

[embphysics]

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.

 

[Curious3141]

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.

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).