Velocity vs time graph simple harmonic motion phase constant

In summary, the given question asks to find the phase constant for a given graph and equation involving displacement and velocity. The solution involves calculating the angular frequency and amplitude, and then using the given equation to determine the phase constant. After some calculation, it is determined that the correct phase angle is either π/6 or 5π/6. Another way to determine the correct phase angle is by considering the acceleration at t=0, which must be positive.
  • #1
firezap
29
0

Homework Statement


http://i.imgur.com/u8vUv5a.jpg
Find the phase constant.

Homework Equations


x(t)=Acos(ωt + Φ)
v(t)=-Aωsin(ωt + Φ)
Vmax = ωA
ω=2π/T

The Attempt at a Solution


ω = 2π/12 = 0.5236
A = 60/0.5236 = 114.59 cm
v(0) = -30 = -114.59(0.5236)sinΦ
0.5 = sinΦ
Φ = π/6 and 5π/6. Which angle is correct?
 
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  • #2
firezap said:
ω = 2π/12
T ≠ 12
Φ = π/6 and 5π/6. Which angle is correct?

There are different ways to see which is correct. For example, you could sketch a graph of V(t) for Φ = π/6 and for Φ = 5π/6 and compare your sketches with the given graph. Or, you could evaluate V(t) for some particular value of t (not equal to zero), say t = 1, for Φ = π/6 and for Φ = 5π/6 and compare to the given graph.
 
  • #3
TSny said:
T ≠ 12
What's the period and how to find it?
So my period and amplitude and angles are incorrect?
 
Last edited:
  • #4
TSny said:
T ≠ 12
Looks like 12 to me.
 
  • #5
thanks, 5π/6 was correct. Confirmed by mastering physics.
 
  • #6
haruspex said:
Looks like 12 to me.
Must be my eyesoo) (or my brain o0)) But it doesn't look like the graph of V(t) goes through exactly one cycle in 12 seconds.

[Oh wait! Now that I look at it again for the 7th time, I see that the graph doesn't meet the t axis at t = 12. Very sorry about that. I agree that T = 12.]
 
  • #7
firezap said:

Homework Statement


http://i.imgur.com/u8vUv5a.jpg
Find the phase constant.

Homework Equations


x(t)=Acos(ωt + Φ)
v(t)=-Aωsin(ωt + Φ)

The Attempt at a Solution


0.5 = sinΦ
Φ = π/6 and 5π/6. Which angle is correct?

The other way to choose the correct phase angle:
The v(t) graph is increasing at t=0. The slope of the v(t) graph is the acceleration, a(t) = dv/dt = -Aω2cos(ωt + Φ). It has to be positive at t=0. Which phase angle to choose?
 
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Related to Velocity vs time graph simple harmonic motion phase constant

1. What is a velocity vs time graph for simple harmonic motion?

A velocity vs time graph for simple harmonic motion is a graph that shows the relationship between the velocity of an object and time during its oscillatory motion. The velocity is typically plotted on the y-axis and time on the x-axis.

2. How can you identify simple harmonic motion from a velocity vs time graph?

Simple harmonic motion can be identified from a velocity vs time graph by observing a sinusoidal curve, where the velocity changes direction at regular intervals and has a constant amplitude. The graph also shows that the velocity is directly proportional to the displacement from the equilibrium position.

3. What does the phase constant represent in a velocity vs time graph for simple harmonic motion?

The phase constant in a velocity vs time graph for simple harmonic motion represents the initial phase or starting point of the oscillation. It is the value of time at which the object starts its motion and it determines the position of the object at any given time during its oscillation.

4. How does the phase constant affect the shape of a velocity vs time graph for simple harmonic motion?

The phase constant affects the shape of a velocity vs time graph for simple harmonic motion by determining the position of the object at any given time. A change in the phase constant will result in a shift of the entire graph, but the shape and amplitude of the curve will remain the same.

5. Can the phase constant be negative in a velocity vs time graph for simple harmonic motion?

Yes, the phase constant can be negative in a velocity vs time graph for simple harmonic motion. A negative phase constant indicates that the object is starting its motion at a position below the equilibrium point, and the velocity curve will be shifted downwards from the x-axis.

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