Understanding Simple Harmonic Oscillators: Phase Constant Explained

[Rave Grrl]

Can someone explain this:

For question A I originally got around .142 M, but that was apparently wrong, because I assumed the phase constant was zero. Can someone explain what the phase constant is and how to find it?


A simple harmonic oscillator consists of a block of mass 2.60 kg attached to a spring of spring constant 180 N/m. When t = 3.00 s, the position and velocity of the block are x = 0.129 m and v = 3.415 m/s.
(a) What is the amplitude of the oscillations?

(b) What was the position of the mass at t = 0 s?

(c) What was the velocity of the mass at t = 0 s?

 

[Sirus]

http://www.kineticbooks.com/physics/16618/27251/sp.html 's a link for phase constant.

Use energy concepts to solve this problem. Remember that mechanical energy is conserved, so the sum of the kinetic and elastic potential energies of the mass is constant throughout its movement. Energy gradually transfers between the two types as the mass moves. Here are some useful formulae:

$$E_{\mbox{k}}=\frac{1}{2}mv^{2}$$

$$E_{\mbox{elastic potential}}=\frac{1}{2}kx^{2}$$

where k is the spring constant and x is the distance from equilibrium of the mass.

 

[thepatientmental]

The phase constant, also known as the phase angle, is a measure of the initial position and velocity of the oscillator. It represents the starting point of the oscillations on the oscillation graph. In simple harmonic motion, the phase constant is the angle between the maximum displacement and the initial position of the oscillator. It is usually denoted by the Greek letter "phi" (ϕ).

To find the phase constant, we first need to understand that the equation for simple harmonic motion is given by x(t) = A cos(ωt + ϕ), where A is the amplitude, ω is the angular frequency, and ϕ is the phase constant.

(a) To find the amplitude, we can use the given information of the position and velocity at t = 3.00 s. We know that at this time, the position is x = 0.129 m, and the velocity is v = 3.415 m/s. Using the equation for simple harmonic motion, we can plug in these values and solve for the amplitude A.

x(t) = A cos(ωt + ϕ)
0.129 m = A cos(ω(3.00 s) + ϕ)
3.415 m/s = -ωA sin(ω(3.00 s) + ϕ)

Solving for A, we get A = 0.129 m. This is the amplitude of the oscillations.

(b) To find the initial position of the mass at t = 0 s, we can use the given information of the position and velocity at t = 3.00 s and the amplitude we just found.

x(t) = A cos(ωt + ϕ)
0.129 m = 0.129 m cos(ω(3.00 s) + ϕ)
cos(ω(3.00 s) + ϕ) = 1

Since the cosine of any angle is equal to 1, we can conclude that ω(3.00 s) + ϕ = 0. Therefore, the initial position of the mass at t = 0 s is x = A = 0.129 m.

(c) To find the initial velocity of the mass at t = 0 s, we can use the given information of the position and velocity at t = 3.00 s and the amplitude we just found.

x(t) = A cos(ωt