Simple Harmonic motion

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Homework Statement


A pendulum of length 2.0 m makes small angle oscillations with an amplitude of 15 degrees.
a) Find the time required for the bob to oscillate from 5 degrees to 10 degrees to the right.
b)Calculate the velocity and acceleration at these two positions.

Homework Equations


ω^2 =g/l ,for small angles [not sure]


The Attempt at a Solution


I'm stuck from a ,so I only attempted a part.
ω = (10/2)^(1/2)
t = 5(pi)/180 / ω = 5(pi)/180 / (10/2)^(1/2) = 0.039 s
The answer is 0.18s which is different from mine significantly.
 

Answers and Replies

  • #2
jambaugh
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If I were presented with this problem, I would for part a.) write down the position and velocity functions as sinusoidal functions of time and work from there. For part b.) I would look at conservation of energy with kinetic and potential components.
 
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  • #3
haruspex
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t = 5(pi)/180 / ω
In the subject of pendulums undergoing SHM it is easy to confuse two things, as you have.
In the expression y=A sin(ωt), as applied to a pendulum, the A and y represent real angles, namely, the maximum and instantaneous angles of swing, respectively. The ωt term does not represent a physical angle. It represents a 'phase' angle, i.e. it tells you the phase of SHM the pendulum is in at time t.
 
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  • #4
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In the subject of pendulums undergoing SHM it is easy to confuse two things, as you have.
In the expression v), as applied to a pendulum, the A and y represent real angles, namely, the maximum and instantaneous angles of swing, respectively. The ωt term does not represent a physical angle. It represents a 'phase' angle, i.e. it tells you the phase of SHM the pendulum is in at time t.
For part a , I did it as
5 = 15 sin (root5 t1)
t1=0.152
10 = 15 sin (root5 t2)
t2=0.326
t2-t1=0.174 s
the answer is close but still wrong. Can you point out where's the mistake? I'm not familiar with the concept, sorry for not being familiar with SHM.

For part b, the answers are 1.09m/s ,0.87m/s^2 at 5 degree and 0.86m/s, 1.74m/s^2 at 10 degree
I found the velocity by conservation of energy, and found acceleration by differentiating -A cos(ωt) (take left as negative) two times and get a=w^2(x).

But I cant find the velocity by differentiating x=-A cos(ωt) and get v = 0.52 root 5 sin(ωt), then v=0.52 root5 sin(20/60 x 2pi) = 1.01m/s which has significant error to the correct answer.

So, I hope to know where my attempt on velocity by differentiation of part b goes wrong.

Also, isn't energy scalar? So if I use energy conservation, the velocity I found is actually speed, but by differentiation, the acceleration I got is only along the x- axis, so I feel contradiction on my answer.
 
  • #5
haruspex
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I cant find the velocity by differentiating x=-A cos(ωt)
I'm not sure what you mean here. In part a, you took the equation of SHM to be x=A sin(ωt). How did you get this cos equation? Is that before or after differentiating?
 
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  • #6
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I'm not sure what you mean here. In part a, you took the equation of SHM to be x=A sin(ωt). How did you get this cos equation? Is that before or after differentiating?
for the phase, I think it should be started from 0, so I took sin in part a, but for distance, I think it starts from -A if I take the center as 0, so I take cos?
 
  • #7
jambaugh
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Right, its just a matter of choosing when t = 0 and this choice of phase/starting time is immaterial to the problem. Imagine (or actually draw) a graph of the function, pick the correspoinding points where the y-value (which haruspex has pointed out is your actual pendulum angle) is at the 5 and then 10 degree marks. Solve for the t's with inverse trig and take the difference to get the duration. Part a.) is done!
 
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  • #8
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Right, its just a matter of choosing when t = 0 and this choice of phase/starting time is immaterial to the problem. Imagine (or actually draw) a graph of the function, pick the correspoinding points where the y-value (which haruspex has pointed out is your actual pendulum angle) is at the 5 and then 10 degree marks. Solve for the t's with inverse trig and take the difference to get the duration. Part a.) is done!
Thanks for your teachings! but as I did that on the upper comment I found 0.174s.... not 0.18s...
 
  • #9
haruspex
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t2-t1=0.174 s
the answer is close but still wrong
If you take g as 9.8 m/s2 instead of 10, then round the answer to two sig figs, you will get .18.
for the phase, I think it should be started from 0, so I took sin in part a, but for distance, I think it starts from -A
In using Asin(ωt) you were defining t=0 as the time when displacement is zero, i.e. the pendulum is vertical. There is no need to change from that for part b. Show your working from there.
 
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  • #10
SammyS
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Thanks for your teachings! but as I did that on the upper comment I found 0.174s.... not 0.18s...
I get t10 - t5 ≈ 0.1763 seconds when using g = 9.8 m/s2 .

When using g = 10 m/s2, I get about 0.1744 seconds .
 
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  • #11
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If you take g as 9.8 m/s2 instead of 10, then round the answer to two sig figs, you will get .18.

In using Asin(ωt) you were defining t=0 as the time when displacement is zero, i.e. the pendulum is vertical. There is no need to change from that for part b. Show your working from there.

I now use radians.
the largest amplitude can be A sin (15π/180)= 0.518m
x=Asin(ωt)
v=Aωcos(ωt)
=0.518 (root5) cos( 5/60 x 2π ) = 1.00 m/s for 5º
v=0.518 (root5) cos( 10/60 x 2π ) =0.580 m/s for 15º

I don't understand why the suggested answers are in scalar.
 
  • #12
haruspex
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I now use radians.
the largest amplitude can be A sin (15π/180)= 0.518m
x=Asin(ωt)
v=Aωcos(ωt)
=0.518 (root5) cos( 5/60 x 2π ) = 1.00 m/s for 5º
v=0.518 (root5) cos( 10/60 x 2π ) =0.580 m/s for 15º

I don't understand why the suggested answers are in scalar.
You are making the same confusion as you did originally. The angles inside the trig functions are phase angles, not pendulum angles.
 
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  • #13
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You are making the same confusion as you did originally. The angles inside the trig functions are phase angles, not pendulum angles.
I'm sorry, I know they should be phase angles ,so I divided 5 by 60 and see its fraction out of 2pi a period? how to change it to phase angles then? If it is pendulum angles, It is 5pi/180,but I suppose 5/60x2pi is phase angles?
 
  • #14
haruspex
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I'm sorry, I know they should be phase angles ,so I divided 5 by 60 and see its fraction out of 2pi a period? how to change it to phase angles then? If it is pendulum angles, It is 5pi/180,but I suppose 5/60x2pi is phase angles?
I don't know where the 60 comes from, but the 5 is a pendulum angle, and you cannot convert it to a phase angle by multiplication.
It is not the same as the difference between degrees and radians.

A phase angle looks like ωt, or ωt+φ, etc. Your first equation in post 4 is of the form "swing angle at time t = max swing angle x sine(phase angle at time t)".
 
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  • #15
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I don't know where the 60 comes from, but the 5 is a pendulum angle, and you cannot convert it to a phase angle by multiplication.
It is not the same as the difference between degrees and radians.

A phase angle looks like ωt, or ωt+φ, etc. Your first equation in post 4 is of the form "swing angle at time t = max swing angle x sine(phase angle at time t)".
Because the amplitude is 15 degrees, so15x4= 60 degrees is one loop. so 5/60 for 5 degrees is = 1/12 of a cycle.

Using another method to get the answers, I put t1 and t2 from a part and got the correct answers.
 
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  • #16
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Because the amplitude is 15 degrees, so15x4= 60 degrees is one loop. so 5/60 for 5 degrees is = 1/12 of a cycle.

Using another method to get the answers, I put t1 and t2 from a part and got the correct answers.
So I suppose the approach by using energy conservation is not appropriate , as the question is asking for velocity on the x - axis. The answer by energy conservation is 0.869 and 1.10, which using the method of differentiating Asin(wt) is 0.863 and 1.09.
 
  • #17
haruspex
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5 degrees is = 1/12 of a cycle.
It's 1/12 of a cycle in regards to distance traversed, but it's not 1/12 of a cycle in respect of time, so you cannot multiply that by the frequency to get the phase angle.
I put t1 and t2 from a part and got the correct answers.
That is the correct way.
 
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  • #18
SammyS
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So I suppose the approach by using energy conservation is not appropriate , as the question is asking for velocity on the x - axis. The answer by energy conservation is 0.869 and 1.10, which using the method of differentiating Asin(ωt) is 0.863 and 1.09.
This problem certainly can be solved using conservation of energy. In fact, this can be done either directly, without regard to SHM, or done using SHM.
 
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