Oscillations: Pendulum with initial velocity


by Stealth849
Tags: initial, oscillations, pendulum, velocity
Stealth849
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#1
Jan16-13, 09:01 PM
P: 38
1. The problem statement, all variables and given/known data

A child on a 4m long swing is pulled back 1m from the vertical and released with a push imparting a speed of 2m/s. Find an expression for the angle θ, as a function of time, identifying the frequency ω, amplitude θi, and phase constant δ.

2. Relevant equations

θ = θmax*cos(ωt + δ)
ω = √g/l

3. The attempt at a solution

I got a value for omega, from √g/l, of 1.565

The fact that the child starts with initial velocity is really throwing me off. In order to find the phase constant, I've tried looking at equations

v = -ωθ*sin(δ)
θ = θi*cos(δ)

and solving for tan(δ) to get a value for δ of 3.17rad...

I don't even know if this is valid, but I otherwise don't know where to start.

I know that if the child is starting from an angle of 0.253rad with initial velocity, he should travel passed the point where I can use small angle approximation.

I thought also about differentiating the basic equation for harmonic motion θ = θmax*cos(ωt + δ) to get velocity, and setting that velocity equal to 2 at time 0, but there are still two unknowns. Any help will be appreciated. Thanks!
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ehild
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#2
Jan16-13, 11:36 PM
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Quote Quote by Stealth849 View Post
1. The problem statement, all variables and given/known data

A child on a 4m long swing is pulled back 1m from the vertical and released with a push imparting a speed of 2m/s. Find an expression for the angle θ, as a function of time, identifying the frequency ω, amplitude θi, and phase constant δ.

2. Relevant equations

θ = θmax*cos(ωt + δ)
ω = √g/l

3. The attempt at a solution

I got a value for omega, from √g/l, of 1.565

The fact that the child starts with initial velocity is really throwing me off. In order to find the phase constant, I've tried looking at equations

v = -ωθ*sin(δ)
θ = θi*cos(δ)


and solving for tan(δ) to get a value for δ of 3.17rad...

I don't even know if this is valid, but I otherwise don't know where to start.
Your equations are not correct. In the small-angle approximation,

θ(t)=Acos(ωt+δ)

If Ω is the angular velocity of the pendulum, Ω=V/L

Ω(t)=dθ/dt=-Aωsin(ωt+δ)


You push the child inward, so the direction of the initial velocity V is opposite to the displacement theta. Taking V negative you get a phase constant less than pi/2.

The If you can apply small angle approximation depends on the maximum angular displacement, the amplitude. Get A by from the equation A2=θ(0)2+(Ω(0)/ω)2.

ehild
Stealth849
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#3
Jan17-13, 12:01 AM
P: 38
I'm not sure I completely understand..

I am using θi to signify amplitude for a simple pendulum in the equations that you mentioned are incorrect, but when I look at how you say to find amplitude, θ(0) and dθ/dt (0) give me the equations you say are incorrect.

θ(t)=Acos(ωt+δ)

say I am using this, where A is amplitude instead of θi, if t = 0,

θ(t)=Acos(δ)

and if t = 0 here,

Ω(t)=dθ/dt=-Aωsin(ωt+δ)

then

dθ/dt=-Aωsin(δ)

And from those, would I not still need a phase shift value to solve the equation for A in

A^2= A^2cos(δ)^2 - A^2sin(δ)^2

?

Or am I just missing something blindingly obvious...?

ehild
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#4
Jan17-13, 12:31 AM
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Oscillations: Pendulum with initial velocity


Quote Quote by Stealth849 View Post
I'm not sure I completely understand..

I am using θi to signify amplitude for a simple pendulum in the equations that you mentioned are incorrect, but when I look at how you say to find amplitude, θ(0) and dθ/dt (0) give me the equations you say are incorrect.

θ(t)=Acos(ωt+δ)

say I am using this, where A is amplitude instead of θi, if t = 0,

θ(t)=Acos(δ)

and if t = 0 here,

Ω(t)=dθ/dt=-Aωsin(ωt+δ)

then

dθ/dt=-Aωsin(δ)

And from those, would I not still need a phase shift value to solve the equation for A in

A^2= A^2cos(δ)^2 - A^2sin(δ)^2
That is wrong. A^2= A^2cos(δ)^2 +A^2sin(δ)^2, as cos(δ)^2 +sin(δ)^2=1.

Note that the initial θ is not the amplitude. θ(0) = 0.253 rad, as you have found already and dθ/dt is initially -2(m/s)/4(m).


ehild
Stealth849
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#5
Jan17-13, 07:22 AM
P: 38
Okay, I'm starting to see it...

A^2= A^2cos(δ)^2 +A^2sin(δ)^2

where Acos(δ) = 0.253 = θ(0)
and -Aω*sin(δ) = -0.5 = dθ/dt at 0

so

A^2 = (0.253)^2 - 0.5^2

and solve for A.

I don't see how I would find the phase shift however. Does it involve dividing the sin and cos functions and solving for tan (δ)?
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#6
Jan17-13, 08:27 AM
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Quote Quote by Stealth849 View Post
Okay, I'm starting to see it...

A^2= A^2cos(δ)^2 +A^2sin(δ)^2

where Acos(δ) = 0.253 = θ(0)
and -Aω*sin(δ) = -0.5 = dθ/dt at 0

so

A^2 = (0.253)^2 - 0.5^2

and solve for A.
Check the eq. in red, it is not correct.
Quote Quote by Stealth849 View Post
I don't see how I would find the phase shift however. Does it involve dividing the sin and cos functions and solving for tan (δ)?
Yes, you do just that.

ehild
Stealth849
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#7
Jan17-13, 10:48 AM
P: 38
I have to divide the second term by ω = 1.565, yes?

A^2 = (0.253)^2 - (0.5/1.565)^2

Then

tan (δ) = AΩ(0)/-Aωθ(0)

I don't know how to determine which quadrant the function would be in, as in how to determine how sin and cos are positive or negative. Would only cos be negative if the denominator is negative in the equation?
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#8
Jan17-13, 02:36 PM
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Quote Quote by Stealth849 View Post
I have to divide the second term by ω = 1.565, yes?

A^2 = (0.253)^2 - (0.5/1.565)^2
Why minus???????

Quote Quote by Stealth849 View Post
Then

tan (δ) = AΩ(0)/-Aωθ(0)

I don't know how to determine which quadrant the function would be in, as in how to determine how sin and cos are positive or negative. Would only cos be negative if the denominator is negative in the equation?

What is Ω(0) = V/L=-2/4. So both sin(δ) and cos(δ) is positive, δ is in the first quadrant.

In the second quadrant, sine is + and cosine is -
In the third quadrant, bot sine and cosine are -
In the fourth quadrant, sine is - and cosine is +

ehild
Stealth849
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#9
Jan17-13, 02:58 PM
P: 38
Ahh, forgot the fact that it is squared. Sorry that took so long to catch. I see it is positive now, not negative.

So ultimately, we have an equation where

A = 0.408

tan(δ) = -0.204/-0.162 = 0.33

in quadrant 1,

δ = 0.033

so

θ(t) = 0.408*cos(1.565t + 0.033) ?
ehild
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#10
Jan17-13, 11:32 PM
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Quote Quote by Stealth849 View Post
Ahh, forgot the fact that it is squared. Sorry that took so long to catch. I see it is positive now, not negative.

So ultimately, we have an equation where

A = 0.408

tan(δ) = -0.204/-0.162 = 0.33?????

in quadrant 1,

δ = 0.033????

so

θ(t) = 0.408*cos(1.565t + 0.033) ?

The amplitude is OK. I do not understand how you calculated delta. It is not right.

Acosδ=0.253
-Asinδ=-0.319


Where did -0.204 and -0.162 come from?

ehild
Stealth849
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#11
Jan18-13, 02:26 AM
P: 38
tan (δ) = AΩ(0)/-Aωθ(0)

I multiplied A by Ω(0)

0.408*(-0.5)

and -A by ωθ(0)

-0.408*1.565*2.53 = -0.162


I see now I already had the values for Acos(δ) and -Asin(δ)

tan(δ) = 0.319/0.253

δ = 0.9

so my equation for tan, tan (δ) = AΩ(0)/-Aωθ(0) is actually incorrect, because I am changing the ratios if i choose to multiply by reciprocal... I needed to keep the divisions separate to keep the sine and cos ratios. silly silly mistake. sorry for all the trouble.

should be

tan (δ) = (Ω(0)/-Aω)/((θ(0)/A)
ehild
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#12
Jan18-13, 02:40 AM
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It is correct now. No need to divide by A, it cancels when you calculate the tangent.

ehild


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