SHM (finding the phase constant)

  • Thread starter MarcL
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
ehild
Homework Helper
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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
Why do you use the minus sign?

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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