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

In summary, the conversation discusses a homework problem involving a spring connected to a pendulum holding a ball with mass m, length l, spring constant k, spring deflection x, angle of pendulum alpha, and angular velocity w. The goal is to derive the angular acceleration, which is found to be -w^2*sin(alpha). There is some confusion regarding the use of the symbols w and alpha, but it is determined that the correct equation is -w^2*sin(w*t+alpha). However, there is still some uncertainty about the role of alpha in the equation.
  • #1
Xavicamps
12
0

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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  • #2
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 ... .?
 
  • #3
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
 
  • #4
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
 
  • #5
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?
 
  • #6
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
 
  • #7
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.
 
  • #8
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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  • #9
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
 

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

1. How does a pendulum and spring system work?

A pendulum and spring system works by utilizing the force of gravity to create a back-and-forth motion. The pendulum is attached to a spring, which helps to regulate the pendulum's oscillations and keep it in motion.

2. What factors affect the motion of a pendulum and spring system?

The motion of a pendulum and spring system can be affected by several factors, including the length of the pendulum, the strength of the spring, and the force of gravity. Other factors such as air resistance and friction can also impact the system.

3. How do you calculate the period of a pendulum and spring system?

The period of a pendulum and spring system can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the pendulum, and k is the spring constant. This equation assumes that the angle of displacement is small (less than 15 degrees).

4. What is the relationship between the length of the pendulum and its period?

The length of the pendulum and its period have an inverse relationship. This means that as the length of the pendulum increases, the period increases as well. This relationship can be demonstrated by adjusting the length of a pendulum and observing the change in its period.

5. How do you know if your results for a pendulum and spring system are correct?

To check if your results for a pendulum and spring system are correct, you can compare them to known values or use mathematical equations to verify the accuracy of your calculations. Additionally, conducting multiple trials and averaging the results can help to ensure accuracy.

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