Why is B equal to RsinΦ in spring vibration equations?

[Amadeo]

TL;DR

Derivation of Equation of Vibration

For undamped free vibrations, we have the following differential equation.

mu'' + ku = 0

where m is the mass of the object hanging on the end of a spring, and u is the distance from the equilibrium position as a function of time.

This yields u = Acosωt + Bsinωt

where ω is √(k/m) (k=spring constant).

I am having trouble understanding why this can be rewritten as

u = RcosΦcosωt + RsinΦsinωt (which, in turn, = Rcos(ωt -Φ) )

If A represents the initial displacement from equilibrium (ui), I can see how we could set this equal to RcosΦ, (R being the maximum displacement) thereby defining Φ to be that value which makes ui=RcosΦ true. But, I don't see why B must, in that case, necessarily be RsinΦ.

It looks like B must be the initial velocity (vi) multiplied by m/k.

 

[Amadeo]

This just occurred to me:

Since the solution u = Acosωt + Bsinωt is general, A and B can be any two constants. If we choose two arbitrary constants for A and B, this will determine the initial displacement and the initial velocity.

Alternatively, if we choose two arbitrary values for the initial displacement and velocity, these will determine A and B.

In this case, we are choosing values for A and B. These values then determine the initial velocity and displacement. We could choose any values we wish, but we chose these values (A= RcosΦ and B=RsinΦ) because they enable the convenient reformulation Rcos(ωt -Φ).

Any more insights would be appreciated.

 

[GlenLaRocca]

You mentioned a mass hanging from a spring and its displacement (m,k,u). The solution is Acosωt + Bsinωt where both A and B have displacement units. I understand those. The R and Φ look like polar coordinates, range and angle. Does your spring hang from a pivot? Describe exactly what is vibrating.

 

[PeroK]

This just occurred to me:

Since the solution u = Acosωt + Bsinωt is general, A and B can be any two constants. If we choose two arbitrary constants for A and B, this will determine the initial displacement and the initial velocity.

Alternatively, if we choose two arbitrary values for the initial displacement and velocity, these will determine A and B.

In this case, we are choosing values for A and B. These values then determine the initial velocity and displacement. We could choose any values we wish, but we chose these values (A= RcosΦ and B=RsinΦ) because they enable the convenient reformulation Rcos(ωt -Φ).

Any more insights would be appreciated.

Another way to look at it: the set $(A, B)$ is just the set of any two numbers. You could think of this as points in a plane expressed in Cartesian coordinates. Equally, you could express this in polar coordinates $(R\cos \phi, R \sin \phi)$, where $R^2 = A^2 + B^2$ etc.