What is the angular frequency of oscillation?

[hidemi]

Homework Statement

At the instant its angular displacement is 0.32 rad, the angular acceleration of a physical pendulum is -630 rad/s2. What is its angular frequency of oscillation?
A) 6.6 rad/s
B) 14 rad/s
C) 20 rad/s
D) 44 rad/s
E) 200 rad/s

The answer is D

Relevant Equations

ω^2 = (ωo)^2 + 2αθ

ω^2 - (ωo)^2
= 2 (-630) (0.32)
= -403.2

This is what I have now and I stuck here.

 

[Steve4Physics]

Homework Statement:: At the instant its angular displacement is 0.32 rad, the angular acceleration of a physical pendulum is -630 rad/s2. What is its angular frequency of oscillation?
A) 6.6 rad/s
B) 14 rad/s
C) 20 rad/s
D) 44 rad/s
E) 200 rad/s

The answer is D
Relevant Equations:: ω^2 = (ωo)^2 + 2αθ

ω^2 - (ωo)^2
= 2 (-630) (0.32)
= -403.2

This is what I have now and I stuck here.

During simple harmonic motion (SHM), acceleration (α) changes. "ω^2 = (ωo)^2 + 2αθ" is for constant angular accelelation, so can't be used in this question.

For linear SHM in the x-direction say, the essential relationships (the things that make it simple harmonic motion) are that:
- the magnitude of acceleration (a) is proportional to the magnitude of displacement (x) and
- acceleration acts in the opposite direction to the displacement.

This can be expressed as: a = -kx where k is a positive constant.. Note that a and x are functions of time, a(t) and x(t). With some calculus you can show k = ω² where ω is the angular frequency. So we can write
a = -ω²x
This is a key formula you need to understand/learn.

For rotational SHM (e.g. thinking about a pendulum in terms of angular changes) the equivalent formula is
α = -ω²θ
(It's basically the same formula as "a = -ω²x" but α(t) is angular acceleration and θ(t) is angular displacement.)

If you use this formula you can find ω.

 

[hidemi]

During simple harmonic motion (SHM), acceleration (α) changes. "ω^2 = (ωo)^2 + 2αθ" is for constant angular accelelation, so can't be used in this question.

For linear SHM in the x-direction say, the essential relationships (the things that make it simple harmonic motion) are that:
- the magnitude of acceleration (a) is proportional to the magnitude of displacement (x) and
- acceleration acts in the opposite direction to the displacement.

This can be expressed as: a = -kx where k is a positive constant.. Note that a and x are functions of time, a(t) and x(t). With some calculus you can show k = ω² where ω is the angular frequency. So we can write
a = -ω²x
This is a key formula you need to understand/learn.

For rotational SHM (e.g. thinking about a pendulum in terms of angular changes) the equivalent formula is
α = -ω²θ
(It's basically the same formula as "a = -ω²x" but α(t) is angular acceleration and θ(t) is angular displacement.)

If you use this formula you can find ω.

Thank you. I think I got it but just want to double check:

θ = A * sin(ωt)
ω = Aω * cos(ωt)
α = -Aω^2 * sin(ωt)

the last formula is what you were referring, correct?

 

[kuruman]

Thank you. I think I got it but just want to double check:

θ = A * sin(ωt)
ω = Aω * cos(ωt)
α = -Aω^2 * sin(ωt)

the last formula is what you were referring, correct?

Yes, use the last equation. It's a straightforward substitution.

 

[hidemi]

Yes, use the last equation. It's a straightforward substitution.

Thank you so much.

 

[Steve4Physics]

Thank you. I think I got it but just want to double check:

θ = A * sin(ωt)
ω = Aω * cos(ωt)
α = -Aω^2 * sin(ωt)

the last formula is what you were referring, correct?

Hi @hidemi. I'm not sure if there is some confusion here, so I'm adding this. The three formulae you list are:
θ = A sin(ωt)
ω = Aω cos(ωt)
α = -Aω² sin(ωt)

You cannot use the last formula directly, because you do not know A or t.

However note that if you substitute "θ = A sin(ωt)" (the first formula) into the last formula, you can easily show that α = -ω²θ. And "α = -ω²θ" is the formula you need to solve this problem.

 

[hidemi]

Hi @hidemi. I'm not sure if there is some confusion here, so I'm adding this. The three formulae you list are:
θ = A sin(ωt)
ω = Aω cos(ωt)
α = -Aω² sin(ωt)

You cannot use the last formula directly, because you do not know A or t.

However note that if you substitute "θ = A sin(ωt)" (the first formula) into the last formula, you can easily show that α = -ω²θ. And "α = -ω²θ" is the formula you need to solve this problem.

Thanks for the clarification!