What Are the Angular Frequency and Amplitude of a Particle Undergoing SHM?

In summary: You end up with the same thing you have above, but with a bit less work.In summary, we can find the angular frequency ω and amplitude A in terms of the given quantities of velocity and displacement using the equations for simple harmonic motion. By eliminating time and initial phase, we can obtain expressions for ω and A in terms of v1, v2, x1, and x2, allowing us to solve for the unknown quantities.
  • #1
NewtonsHead
26
0

Homework Statement



A particle undergoes simple harmonic motion. It has velocity v1 when the displacement is x1 and velocity v2 when the displacement is x2. Find the angular frequency ω and amplitude A in terms of the given quantities.

Homework Equations



x = A sin (ωt + ∅ )

v = A ω cos ( ωt + ∅ )

The Attempt at a Solution



I tried starting x1 and v1 at t=0 s. This yields

x = A sin ( ∅ )

v = A ω cos ( ∅ )
The equations for position 2 included the ωt. I have four equations: two position and two velocity and four unknowns: A, ω, ∅, and t. I just need A and ω. Is this the right direction?
 
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  • #2
NewtonsHead said:
x = A sin (ωt + ∅ )

v = A ω cos ( ωt + ∅ )

... I have four equations: two position and two velocity and four unknowns: A, ω, ∅, and t. I just need A and ω. Is this the right direction?

Welcome to PF! Your approach is fine. (Another approach is to consider energy relations.) What do you get if you divide your velocity equation above by ω and then square the equation? How does that compare with squaring the x equation? Can you see how to eliminate ∅ and t in one fell swoop?
 
  • #3
x2 = A2 sin2 (ωt + ∅ )

(v/ω)2 = A2 cos2 ( ωt + ∅ )

I could say

A = √ (v/ω)2 + x2

I don't see how I can eliminate t and ∅ though
 
  • #4
NewtonsHead said:
x2 = A2 sin2 (ωt + ∅ )

(v/ω)2 = A2 cos2 ( ωt + ∅ )

I could say

A = √ (v/ω)2 + x2
Right. And you have two such equations, one for x1 and v1, and one for x2 and v2.
I don't see how I can eliminate t and ∅ though
You just did. :biggrin:
 
  • #5
Okay guys, I believe I have the right answer.

I solved for A in each set of x and v.

A = √ (v1/ω)2 + (x1)2 and A = √ (v2/ω)2 + (x2)2

Then I solved for ω in the second equation ( 2 )

I substituted this into the first equation. After a bunch of algebra, I obtained an answer in terms of the given values which makes me happy.

A = √ [ ( (x1)2 (v2)2) ) - ( (x2)2) (v1)2) / ( (v2)2) - (v1)2 ) ]

Then I just had to substitute this value for A in equation 1 to find ω.

Thanks for the help TSny and haruspex
 
  • #6
NewtonsHead said:
I solved for A in each set of x and v.

A = √ (v1/ω)2 + (x1)2 and A = √ (v2/ω)2 + (x2)2

Then I solved for ω in the second equation ( 2 )

I substituted this into the first equation. After a bunch of algebra, I obtained an answer in terms of the given values which makes me happy.

A = √ [ ( (x1)2 (v2)2) ) - ( (x2)2) (v1)2) / ( (v2)2) - (v1)2 ) ]

Then I just had to substitute this value for A in equation 1 to find ω.

Looks very good! You can save some effort by not taking the square roots. You have

A2 = (v1/ω)2 + (x1)2

A2 = (v2/ω)2 + (x2)2

Subtracting these two equations should allow you to fairly easily find ω. Then you can find A.
 

1. What is SHM (Simple Harmonic Motion)?

Simple Harmonic Motion (SHM) is a type of periodic motion in which a particle moves back and forth around a stable equilibrium position, with a constant amplitude and period. It is characterized by a restoring force that is directly proportional to the displacement from equilibrium.

2. What are the necessary conditions for a particle to undergo SHM?

The necessary conditions for a particle to undergo SHM are a restoring force that is directly proportional to the displacement from equilibrium, and a stable equilibrium position. Additionally, there should be no external damping or friction forces present.

3. How is the motion of a particle undergoing SHM represented mathematically?

The motion of a particle undergoing SHM can be represented mathematically using the equation x = A cos(ωt + ϕ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, t is time, and ϕ is the phase angle.

4. What is the relationship between the period and frequency of a particle undergoing SHM?

The period (T) and frequency (f) of a particle undergoing SHM are inversely proportional to each other. This means that as the frequency increases, the period decreases, and vice versa. The relationship between them can be represented mathematically as T = 1/f or f = 1/T.

5. How does the amplitude affect the motion of a particle undergoing SHM?

The amplitude (A) of a particle undergoing SHM determines the maximum displacement from equilibrium that the particle will reach. A larger amplitude will result in a wider oscillation of the particle, while a smaller amplitude will result in a narrower oscillation. The amplitude also affects the energy of the system, with a larger amplitude resulting in a higher energy.

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