I Question about the Derivation of the Equations of Vibration

AI Thread Summary
The discussion revolves around the differential equation for undamped free vibrations, represented as mu'' + ku = 0, leading to the general solution u = Acosωt + Bsinωt. The challenge is understanding the transformation of this equation into the form u = Rcos(ωt - Φ), where R and Φ relate to initial conditions. It is clarified that A and B can be any constants, and by selecting A = RcosΦ and B = RsinΦ, one can conveniently reformulate the solution. The relationship between A, B, and the initial displacement and velocity is emphasized, illustrating how these values can be chosen to fit the system's dynamics. Overall, the discussion highlights the flexibility in representing the solution to the vibration equation while maintaining the connection to physical parameters.
Amadeo
Messages
28
Reaction score
9
TL;DR Summary
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.
 
Physics news on Phys.org
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.
 
  • Like
Likes Dale and PeroK
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.
 
Last edited:
Amadeo said:
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.
 
  • Like
Likes nasu and vanhees71
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top