Determining angular frequency and amplitude for SHM

In summary, the conversation discusses finding the angular frequency and amplitude of a mass attached to a spring with a given speed and position. The first part uses trigonometric functions to find the angular frequency, while the second part utilizes conservation of energy to find both the angular frequency and amplitude. The final expressions for ω0 and A are ω0 = sqrt [ (v12 - v22)/(x22 - x12) ] and A = sqrt [ (v1/ω0)^2 + x1^2 ], respectively.
  • #1
Elvis 123456789
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Homework Statement


A mass "m" is attached to a spring of constant "k" and is observed to have an amplitude "A" speed of "v0" as it passes through the origin.

a) What is the angular frequency of the motion in terms of "A" and "v0"?

b) Suppose the system is adjusted so that the mass has speed "v1" at position "x1" and speed "v2" at position "x2". Find the angular frequency, "ω0", and amplitude "A" in terms of these given quantities. Simplify your results.

Homework Equations


x = A*sin(ω0*t -φ)

v = A*ω0*cos(ω0*t -φ)

The Attempt at a Solution


so for the first part

x(t0) = A*sin(ω0*t0 -φ) = 0

therefore φ = ω0*t0

v(t0) = A*ω0*cos(ω0*t0 -φ) = v0

therefore ω0 = v0/A

for part b.)

x1 = A*sin(ω0*Δt1)
v1 = A*ω0*cos(ω0*Δt1)

x2 = A*sin(ω0*Δt2)
v2 = A*ω0*cos(ω0*Δt2)

where Δt1 = t1 - t0
and Δt2 = t2 - t0theres four unknowns and four equations so I know I should be able to solve for ω0 and A, but I can't seem to untangle all these trig functions. any hints?
 
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  • #2
I suggest taking a look at an energy conservation approach. Both the KE and PE expressions square their variable so position and velocity directions don't matter there.
 
  • #3
gneill said:
I suggest taking a look at an energy conservation approach. Both the KE and PE expressions square their variable so position and velocity directions don't matter there.
Ok so using conservation of energy

KE1 + PE1 = KE2 + PE2

1/2*m*v12 + 1/2*k*x12 = 1/2*m*v22 + 1/2*k*x22

canceling the halves and dividing by "m" on both sides

v12 + ω02*x12 = v22 + ω02*x22

then rearranging for ω0
ω0 = sqrt [ (v12 - v22)/(x22 - x12)

for the amplitude I use the fact that the total energy at any time is equal to 1/2*k*A2

so KE1 + PE1 = 1/2*k*A2

plugging in and rearranging I get

A = sqrt [ x12 + v12 * (x22 - x12)/(v12 - v22) ]

does this look okay?
 
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  • #4
That looks good.

For the amplitude you could also have chosen just one of the given (x,v) pairs to find the total energy. The resulting expression might be a bit cleaner...
 
  • #5
gneill said:
That looks good.

For the amplitude you could also have chosen just one of the given (x,v) pairs to find the total energy. The resulting expression might be a bit cleaner...
I don't quite understand what you mean. I chose just x1 and v1 but x2 and v2 factor in because I had to divide by the mass again to get the angular frequency since they didn't want the answer in terms of "k"
 
  • #6
You're right, I missed that stipulation. Mind you, if you find ##\omega_o## first, you should be able to use that in the amplitude expression. You should be able to reach:

##A = \sqrt{\left(\frac{v}{\omega_o}\right)^2 + x^2}##
 
  • #7
gneill said:
You're right, I missed that stipulation. Mind you, if you find ##\omega_o## first, you should be able to use that in the amplitude expression. You should be able to reach:

##A = \sqrt{\left(\frac{v}{\omega_o}\right)^2 + x^2}##
Yes, that's the expression that I arrived at before I plugged in the corresponding values. Thanks for the help :)
 

1. What is the formula for calculating angular frequency in SHM?

The formula for calculating angular frequency in SHM is ω = √(k/m), where ω is the angular frequency, k is the spring constant, and m is the mass of the object undergoing SHM.

2. How do you calculate the amplitude of an object in SHM?

The amplitude of an object in SHM can be calculated by finding the maximum displacement from equilibrium. This can be determined by measuring the distance from the equilibrium position to the highest or lowest point of oscillation.

3. What is the relationship between angular frequency and period of oscillation in SHM?

The relationship between angular frequency and period of oscillation in SHM is T = 2π/ω, where T is the period and ω is the angular frequency. This means that as the angular frequency increases, the period decreases and vice versa.

4. How does changing the spring constant affect the angular frequency in SHM?

Changing the spring constant will directly affect the angular frequency in SHM. A higher spring constant will result in a higher angular frequency, while a lower spring constant will result in a lower angular frequency.

5. Can the amplitude of SHM change over time?

No, the amplitude in SHM remains constant as long as the system remains undisturbed. It is only affected by external forces such as damping or friction, which can decrease the amplitude over time.

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