Simple harmonic motion pendulum ( only given length and release angle)

In summary: So the equation would be θ(0) = Θ sin Φ.I also get that Φ would be the angle the pendulum starts at when released and Θ would be the angle the pendulum swings to. I get that at t=0, θ is 0.10 rad and ω is 0.
  • #1
Mr Person
4
0
Hi everyone, this is my first post on this forum. I hope you can help and maybe one day I can help some of you ;)

Homework Statement



I need to determine the constants of pendulum. Θ ( in radians) and Φ.

I am given the length of the pendulum, 0.30m

and the initial angle of release, 0.10 rad

and g = 9.81 m s^-2

Homework Equations



I have the equation

θ (t) = Θ sin(ωt +Φ)

The Attempt at a Solution



I know that ω = √(g/l)

so ω = 5.72 to 3 s.f.

at t=0

θ = 0.10 rad

ωt = 0

which leaves me with

0.10 = Θ sinΦ

At which point I am stuck,

It seems there can be any number of combinations of Θ and sinΦ that would equal 0.10.
I have worked out the period of the pendulum

using

T = 2 π √ l/g

T = 1.10 to 3.s.f.

but that doesn't seem to get me anywhere.

Thanks in advance :)
 
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  • #2
Welcome to the forums Mr Person! :smile: I hope you find them helpful! (and at this stage in my career, I find helping fun!)

What this problem basically addresses is what goes on at the initial point (this is sometimes known as the "boundary condition"). At the initial point, what is the kinetic and potential energy?

Note it's initial position, and think: If you drop the pendulum, does it ever have the ability to go to a higher angle? One of your unknowns relates to the maximum amplitude of the motion.

Then: what is it's initial angular speed? You should also be able to write an expression for this based on your expression for angular position, then evaluate this expression at t=0. This will hep you find your other unknown, which relates to phase. (You could also find this from your equation too).

What are your thoughts on these things?
 
  • #3
Thanks for the quick response!

Although I'm sure finding the kinetic and potential energies would be a way to solve this problem, its not been covered in my course yet so I think I'll have to solve this using simple harmonic motion techniques only.


I have worked out the initial displacement using cos

cos 0.1 = opp / 0.3m = 0.3m for initial displacement

I think that 0.3m would be maximum amplitude so if i use

x(0) = A sin(Φ)

0.3 = 0.3 sin(Φ)

1= sin (Φ)

Φ = π/2 radians

Which looks right on a a sin graph as π/2 would be the maximum which seems to work with the model.

If i sub this into the point i got stuck at in the first place I get

0.10 = Θ sin π/2

which gives me Θ = 0.1

although I'm not totally happy with that, it doesn't feel right and I think there must be a better way to work out Θ without relaying on what I have worked out for Φ but I'll move on.

I think the angular velocity should be

v(t) = Aωcos (ωt + Φ )

v(o) = 0.1 x 5.72 cos (5.72 x 0 + π/2)

which gives me 0 because cos π/2 is 0 which seems to make sense if the pendulum is released from rest.

I'm torn between thinking this is all right or that I'm going round in circles :)
 
  • #4
Mr Person:

You seem to be doing great thinking this through (you seem to be finding things are consistent, even if you're going about them circularly). Your phi is the "phase angle"... which in this case (with your sine function for position) would be pi/2, since the problem starts from rest at maximum position... not minimum (where phi would be zero, again just because your are using sine... not cosine... and sine and cosine are off from each other by a phase shift of pi/2).

So let's address your "theta" constant... you've done way to much work here. since you start out at maximum amplitude... that theta would just be your angle.
I.e., take this line from your first post:

Mr Person said:
θ (t) = Θ sin(ωt +Φ)

so... at t=0...
0.10 = Θ sinΦ

So with your phase angle pi/2, you see you have 0.1 radians as theta.

So hmm... what did you learn from all this math?
You got your given 0.1 radians (your given) as theta, and pi/2 as phi (to show you're starting the motion shifted to a cosine function.

I think this problem is trying to familiarize you with this... so hopefully you can really quickly find these constants (just by glancing and quickly inspecting... not by really mathematically cranking). I'd suspect some of these to crop up on tests (although I'm not your instructor :wink:).
 
  • #5
Thanks for the continuing help :),

I'm starting to think I am over thinking this...

I get that at t(0) Θ = θ in this case as the pendulum starts the maximum amplitude. In my equation the amplitude is being measured in radians instead of a pure number like in the normal

θ (t) = A sin(ωt +Φ)

My concern at this point is being able prove these constants when initial conditions aren't stacked in my favor so I was trying to find a better way to prove it.

Must crack on !
 
  • #6
Mr Person said:
Thanks for the continuing help :),

My concern at this point is being able prove these constants when initial conditions aren't stacked in my favor so I was trying to find a better way to prove it.

Must crack on !

When they aren't stacked in your favor, just use BOTH position and velocity equations (or in your case, both the angular position and the angular velocity equations) with your initial conditions... you'll have two equations, and hopefully only two unknowns (in your case amplitude and phase). It still shouldn't take two long to solve the set of equations for your unknowns... it just can't be easily done by inspection (which I think WAS the goal of this problem).
 
  • #7
Thanks so much again :)
 

1. What is a simple harmonic motion pendulum?

A simple harmonic motion pendulum is a system in which a mass (called the bob) is suspended from a fixed point by a string or rod. When the bob is pulled to one side and released, it will swing back and forth in a regular pattern known as simple harmonic motion.

2. How does the length of the pendulum affect its motion?

The length of the pendulum affects the period, or the time it takes for one complete swing. A longer pendulum will have a longer period, meaning it will take more time for the pendulum to swing back and forth. This is because the longer pendulum has a greater distance to cover in each swing, resulting in a slower motion.

3. What is the release angle of a pendulum?

The release angle of a pendulum is the angle at which the bob is released from its starting position. It is usually measured from the vertical position, with an angle of 0 degrees representing a pendulum released at its resting position.

4. Can the release angle affect the motion of the pendulum?

Yes, the release angle can affect the motion of the pendulum. The release angle determines the initial potential energy of the pendulum, which will affect the amplitude (maximum displacement) of the pendulum's swing. A larger release angle will result in a larger amplitude, while a smaller release angle will result in a smaller amplitude.

5. How is simple harmonic motion related to pendulums?

Simple harmonic motion is the regular back and forth motion that a pendulum exhibits. This is because the restoring force (gravity) of the pendulum follows Hooke's Law, which states that the force is directly proportional to the displacement from equilibrium. As the pendulum swings, it continuously cycles between potential energy (at the highest points) and kinetic energy (at the lowest points).

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