Pendulum and spring; Would like to check if the result is correct.

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The discussion revolves around deriving the angular acceleration of a pendulum connected to a spring, with a focus on the correct interpretation of variables like angular velocity and frequency. Participants clarify the relationships between angular displacement, angular velocity, and angular acceleration, emphasizing the need for consistent notation. There is confusion regarding the substitution of variables, particularly between angular velocity and frequency, which leads to errors in the derived equations. Ultimately, the correct expression for angular acceleration is debated, with a consensus that it should reflect the system's oscillatory nature. The conversation highlights the importance of clear definitions and accurate mathematical representation in physics problems.
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Homework Statement


Spring connected to a pendulum holding a ballwith mass m, length l, spring constant k, spring deflection x, angle of pendulum alpha, angular velocity w.

Homework Equations


Derive the angular acceleration

The Attempt at a Solution



I made the three body diagram, and find that -k*x-m*g*sin(alpha)=m*a, from here we know that a=angularacceleration*l, x=l*sin(alpha),alpha=w*t and w=sqrt(k/m+g/l) ---> not sure about this one. And I obtained that angularacceleration=w^2*sin(w*t). Is my derivation correct?
 
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Not sure I have the picture. Is there a simple pendulum with a fixed length rod, and a spring connected to the same mass?
If so, how is the spring arranged... vertically below, horizontal...?
Or is the 'rod' itself the spring? Or ... .?
 
Is a pendulum with a fixed length rod and a spring connected to the same mass, in horizontal position. I will try to picture it.
\
\
\
><><><><● where: \ rod, ● mass and >< spring
 
As I can see the result of my "picture" is not the one I was expecting, assume that the rod goes to the ball, and you will see it clear
 
Xavicamps said:
Is a pendulum with a fixed length rod and a spring connected to the same mass, in horizontal position. I will try to picture it.
\
\
\
><><><><● where: \ rod, ● mass and >< spring
Ok. Are we to take the spring as sufficiently long compared with vertical displacements of the mass that the spring is effectively horizontal always?
 
haruspex said:
Ok. Are we to take the spring as sufficiently long compared with vertical displacements of the mass that the spring is effectively horizontal always?
Yes, it is always effective
 
Xavicamps said:
Yes, it is always effective
I guess you mean always horizontal.
You seem to have used w for two different entities, angular velocity and frequency.
Your answer looks wrong since k does not feature.
Try to write the complete differential equation for the horizontal motion, either in terms of spring displacement, x, or in terms of angular displacement, alpha.
 
Here you have the differential eqution, my problem is if in this case the W is sqrt (g/l+k/m)
 

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Xavicamps said:
Here you have the differential eqution, my problem is if in this case the W is sqrt (g/l+k/m)
Yes, that looks right for the frequency, but as I wrote, you have w standing for two different things, frequency and angular velocity.
Please post your working for your final equation for angular acceleration.

Are you sure you have stated the question correctly, that you are to find the angular acceleration as a function of the angular velocity?
I don't see how you can do that without knowing the amplitude.
 
  • #10
yes, you are right, i made a mistake while saying that w was angular velocity, w is the frequency. Therefore we will obtain that acceleration(angular)=-w^2*sin(alpha), now alpha would be substituted by... angular velocity*time?
 
  • #11
Xavicamps said:
yes, you are right, i made a mistake while saying that w was angular velocity, w is the frequency. Therefore we will obtain that acceleration(angular)=-w^2*sin(alpha), now alpha would be substituted by... angular velocity*time?
If the SHM is given by ##\theta=\Theta\sin(\omega t)## then what are the angular velocity and angular accelerations as functions of t?
How can you use that to write the angular acceleration as a function of angular velocity?
 
  • #12
the angular velocity is Θ*w*cos(w*t) and the angular acceleration is -Θ*w^2*sin(w*t). I think alpha should be substituted by alpha initial + w*t, this is the only solution I can think about
 
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  • #13
Xavicamps said:
the angular velocity is Θ*w*cos(w*t) and the angular acceleration is -Θ*w^2*sin(w*t). I think alpha should be substituted by alpha initial + w*t, this is the only solution I can think about
There is still some confusion here we need to clear up. The problem statement reads:
Xavicamps said:
Spring connected to a pendulum holding a ball with mass m, length l, spring constant k, spring deflection x, angle of pendulum alpha, angular velocity w.
?
It doesn't make any sense to me to change that to "frequency w" since you can and have found the frequency. If you want to use w for frequency then we need to assign another symbol for angular velocity. Call it z.
Yes, you can substitute alpha for the wt. We effectively fixed alpha initial by not putting a phase term in the equation. When alpha is zero the acceleration is zero, so alpha initial is zero.

Edit: that was wrong! alpha is not the same as wt. they are quite different beasts. See later post.
 
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  • #14
Definetely the mistake is mine. we do not have the angular velocity, but the "frequency w". Therefore, I think we arrived to the solution we were looking for: -w^2*sin(w*t+alpha), where we know the angle, and the frequency w. Is that correct?
 
  • #15
Xavicamps said:
Definetely the mistake is mine. we do not have the angular velocity, but the "frequency w". Therefore, I think we arrived to the solution we were looking for: -w^2*sin(w*t+alpha), where we know the angle, and the frequency w. Is that correct?
No, I'm sorry, I was wrong to endorse substituting alpha for wt. (Post edited.)
wt is not an actual angle, it is a theoretical angle representing where the system is in its oscillation.
In the generic SHM equation I quoted in post #11, ##\theta=\Theta \sin(\omega t)##, where does alpha fit?
 
  • #16
I think in our case alpha should be equal to Θsin(ωt), as this formula is telling us what is the position of the system for every time unit.
 
  • #17
Xavicamps said:
I think in our case alpha should be equal to Θsin(ωt), as this formula is telling us what is the position of the system for every time unit.
Right. So can you use that to express the angular acceleration as a function of alpha and w?
 
  • #18
yes, the final result would be -w^2*sin(Θsin(ωt)) where Θ would be the elongation of the system
 
  • #19
Xavicamps said:
yes, the final result would be -w^2*sin(Θsin(ωt)) where Θ would be the elongation of the system
No. How do you get that?
 
  • #20
substitution of alpha by Θsin(ωt), is not what we said it was correct in post #17?
 
  • #21
Xavicamps said:
substitution of alpha by Θsin(ωt), is not what we said it was correct in post #17?
Ok, but what did you substitute it into?
Edit: the substitution is the other way. You want to get rid of the wt part in another equation.
 
  • #22
-w^2*sin(Θsin(ωt)+w*t), unless this is the solution I do not get your point
 

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