SHM (finding the phase constant)

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SUMMARY

The discussion focuses on calculating 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 the oscillations is determined to be 0.500 m using the energy formula E=U+K. The position at t=1.00 s is given as x=0.129 m, and the angular frequency is calculated as w=-7.07 rad/s. The user seeks clarification on determining the correct phase constant, Φ, and whether to use the principal value or its supplementary angle.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with the energy conservation principle in oscillatory systems
  • Knowledge of trigonometric functions and their inverses
  • Ability to manipulate equations involving angular frequency and phase constants
NEXT STEPS
  • Learn how to derive the phase constant in simple harmonic motion
  • Study the implications of using different phase constants in SHM equations
  • Explore the concept of angular frequency and its calculation in oscillatory systems
  • Investigate the use of arccos and its range in determining angles
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators teaching concepts related to simple harmonic oscillators.

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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