Solving for Phase Constant: Simple Harmonic Motion

In summary, the block is at 5 cm with a positive velocity at time t = 0 and its motion is simple harmonic with amplitude 10 cm and period 2 seconds. The phase constant φ should be 60 degrees, as determined by setting t = 0 in the equation x(t) = Acos(wt + φ). The initial position of 5 cm is not a measure in degrees or radians, but rather represents a shift in the phase constant.
  • #1
Glorzifen
25
0

Homework Statement


Another SHM question. Thanks for the help on my first one.

The block is at x0 = +5 cm with a positive velocity V0 at time t = 0. Its motion is SHM with amplitude 10 cm and period 2 seconds. If x(t) = Acos(wt + [tex]\varphi[/tex]), the phase constant [tex]\varphi[/tex] should be:

Homework Equations


x(t) = Acos(wt + [tex]\varphi[/tex])


The Attempt at a Solution


0.05 = 0.10cos([tex]\pi[/tex] + [tex]\varphi[/tex])
[tex]\varphi[/tex] = 60 - [tex]\pi[/tex] (since the 60 is in degrees...do I just use pi radians = 3.14 radians?)
= 57 degrees

I got the pi for wt this way btw:
wT = 2pi
w = 2pi/T = pi (since T = 2)

Not sure if I screwed up that or the radians/degrees thing at the end. Any help would be appreciated.
 
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  • #2
The relevant equation is
x(t) = Acos(wt + φ)
If you put t = 0, then
x(0) = Acos(φ)
 
  • #3
Okay. So it is 60 degrees then. Thanks!
 
  • #4
rl.bhat said:
The relevant equation is
x(t) = Acos(wt + φ)
If you put t = 0, then
x(0) = Acos(φ)

Just one more thing about this actually...my first inclination upon reading the 5cm was to use that as the phase constant. It describes it as starting out from there...I thought it normally started at 0 so in order to 'get it to 5cm' I would need to use the phase constant to represent that. Obviously that is wrong...I'm just not quite sure why...

EDIT: Ah...so we are starting 5cm from 0...but 5cm is not a degree or radian measure...what we're doing is finding how much of a shift 5cm constitutes in terms of a phase constant. Is that right?
 

Related to Solving for Phase Constant: Simple Harmonic Motion

1. What is the phase constant in simple harmonic motion?

The phase constant in simple harmonic motion is a constant that represents the initial position of the object at time t = 0. It determines the starting point of the motion and is denoted by the symbol φ.

2. How is the phase constant related to the amplitude and period of the motion?

The phase constant is related to the amplitude and period of the motion through the equation φ = -arctan(A/B), where A is the amplitude and B is the period. This means that the phase constant is inversely proportional to the amplitude and directly proportional to the period.

3. Can the phase constant change during simple harmonic motion?

No, the phase constant remains constant throughout the entire motion. It only determines the initial position of the object and does not change as the object moves through its cycle.

4. How can the phase constant be calculated using the initial conditions of the system?

The phase constant can be calculated using the equation φ = -arctan(x₀/A), where x₀ is the initial position of the object and A is the amplitude of the motion. This equation can be derived from the equation of motion for simple harmonic motion, x = A cos(ωt + φ), by substituting t = 0 and x = x₀.

5. What is the significance of the phase constant in simple harmonic motion?

The phase constant is important in simple harmonic motion as it determines the initial position of the object and can affect the overall shape and behavior of the motion. It also allows us to calculate the displacement, velocity, and acceleration of the object at any given time during the motion.

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