# Velocity vs time graph simple harmonic motion phase constant

## 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?

TSny
Homework Helper
Gold Member
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.

T ≠ 12
What's the period and how to find it?
So my period and amplitude and angles are incorrect?

Last edited:
haruspex
Homework Helper
Gold Member
T ≠ 12
Looks like 12 to me.

thanks, 5π/6 was correct. Confirmed by mastering physics.

TSny
Homework Helper
Gold Member
Looks like 12 to me.
Must be my eyesoo) (or my brain ) 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.]

ehild
Homework Helper

## 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?

MUUKGIZ