SHM (finding the phase constant)

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The discussion revolves around finding the phase constant for a simple harmonic oscillator with a mass of 2.00 kg and a spring constant of 100 N/m. The amplitude of oscillation is determined to be 0.500 m using the energy formula. The user is attempting to calculate the phase constant using the equation x = Xm cos(wt + Φ), where they have derived a value of 1.31 for Φ but are unsure how to handle the negative angle and whether to use the alternative angle from 2π. Clarification is sought on the correct approach to determine the phase constant within the range of 0 to 2π. The discussion highlights the complexities of phase constant calculations in harmonic motion.
MarcL
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Homework Statement



A simple harmonic oscillator consists of a block of mass 2.00
kg attached to a spring of spring constant 100 N/m.When t =1.00
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? What were
the (b) position and (c) velocity of the block at t =0 s?

I have found a which is Xm = .500m ( I found it through the energy formula E=U+K, solve for x then Xm). But I need my phase constant to do b and c


Homework Equations



x=Xmcos (wt + Φ)


The Attempt at a Solution



I have found my 2 possible phase constant variable doing this

.129 = .500cos(-7.07(1) + Φ) --> I found w using -(k/m)^1/2

I am left with arccos .129/.500 = -7.07 + Φ

And it's here I always get loss... 1.31 = -7.07+ Φ
Do I need to do 2pi - 1.31 to get the second angle? If so, how do I decide which one to use?
 
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MarcL said:

Homework Statement



A simple harmonic oscillator consists of a block of mass 2.00
kg attached to a spring of spring constant 100 N/m.When t =1.00
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? What were
the (b) position and (c) velocity of the block at t =0 s?

I have found a which is Xm = .500m ( I found it through the energy formula E=U+K, solve for x then Xm). But I need my phase constant to do b and c


Homework Equations



x=Xmcos (wt + Φ)


The Attempt at a Solution



I have found my 2 possible phase constant variable doing this

.129 = .500cos(-7.07(1) + Φ) --> I found w using -(k/m)^1/2

Why do you use the minus sign?

MarcL said:
I am left with arccos .129/.500 = -7.07 + Φ

And it's here I always get loss... 1.31 = -7.07+ Φ
Do I need to do 2pi - 1.31 to get the second angle? If so, how do I decide which one to use?

Find a phase constant between 0 and 2pi.

ehild
 
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