Solve SHM Problem: 1 Initial Condition

[JNBirDy]

What is the best way to go about solving a SHM problem that asks for amplitude and phase constant when only one initial condition (x(0)) is given? I'm a little confused because I read that to find A and the phase constant you must use the initial conditions - but I've encounter a few problems which only give one initial condition. I'm obviously missing some understanding... here's an example:

Homework Statement

You observe a vibration that has a frequency of 45 Hz. At t = 0 the displacement is 20 mm and at t = 7 ms the displacement is -12 mm. What is the amplitude (A) and phase constant (⌀) for the equation x(t) = Acos(ωt + ⌀)

Homework Equations

x(t) = Acos(ωt + ⌀)
T = 1/f
ω = 2πf = 2π/T

The Attempt at a Solution

ω = 2πf
ω = 2π(45)
ω = 282.7 rads/s

x(t) = Acos(ωt + ⌀)
x(0) = 0.02 = Acos((282.7)(0) + ⌀)
0.02 = Acos(⌀)
A = 0.02/cos(⌀)

Now what? Do I do the same with x(0.007) even though A and ⌀ are supposed to be found using initial conditions?

Any help is appreciated.

 

[vela]

Those constants don't have to be found using only initial conditions. The important point is that to determine the two unknowns, you need two independent conditions. In this problem, the two pieces of info are x(0)=20 mm and x(7 ms)=-12 mm.

 

[JNBirDy]

Once I have the two equations:

0.02 = Acos(⌀)

-0.012 = Acos(1.98 + ⌀)

How exactly do I manipulate them so that I can solve for one of them? I know that I can get them into the form,

A = 0.02 / cos(⌀)

A = -0.012 / cos(1.98 + ⌀)

and

⌀ = arccos(0.02 / A)

⌀ = arccos(-0.012 / A) - 1.98

but I'm having trouble actually solving it.

 

[vela]

Try using the angle-addition identity for cosine and set the two expressions you have for A equal to each other.

 

[JNBirDy]

Try using the angle-addition identity for cosine and set the two expressions you have for A equal to each other.

Are you referring to:

cos(α+β) = cos(α)cos(β) - sin(α)sin(β)

cos(α-β) = cos(α)cos(β) + sin(α)sin(β)

If so I don't understand how to use them in this case since

0.02 / cos(⌀) = -0.012 / cos(1.98 + ⌀)

cos(1.98 + ⌀) / cos(⌀) = -0.6

----

Maybe this is what you wanted me to do:

cos(1.98 + ⌀) / cos(⌀) = -0.6

cos(1.98)cos(⌀) - sin(1.98)sin(⌀) / cos⌀ = -0.6

cos(1.98) - sin(1.98)tan(⌀) = -0.6

tan⌀ = (-0.6 - cos(1.98))/-sin(1.98)

⌀ = 0.2203 rads

A = 0.02/cos⌀

A = 0.0205 m

?