Finding the Oscillations of Pendulum A

In summary: There might be a phase difference between the pendulum A and B, which would cause the pendulum A to reach the minimum amplitude first.
  • #1
happyparticle
405
20
Homework Statement
Finding when the oscillations of the mass A will be at the minimum amplitude for the first time.

At t = 0, pendulum A = ##x_a = 5cm## and pendulum B = ##x_a = 0##
Relevant Equations
normal modes coordinates
##q_p(t) = A cos \omega_p t##
##q_b(t) = B cos \omega_b t##
First of all, I found the angular frequencies for both pendulum and breathing mode which are
##\omega_p = 4.95##
##\omega_b = 7.45##

Then I found the normal mode coordinates equations:

##q_p(t) = A cos \omega_p t##
##q_b(t) = B cos \omega_b t##

And the beating frequency (I'm not sure if I need it or even if the value is correct)
##f_{beat} = 0.4##

Because the initial velocities are zero, the amplitude of each normal mode is equal to the initial value of the normal mode coordinate.
The pendulum A is at ##x_a = 5## and because the pendulum B is at ##x_a = 0## the initial position of the pendulum a (I guess)##q_p = x_1 + x_2, q_b = x_1 - x_2##
##q_p = 5 + ?, q_b = 5 - ?##

then

##q_p(t) = (5 + ?) cos \omega_p t##
##q_b(t) = (5-?) cos \omega_b t##

I know as well that
##x_1 = \frac{q_p(t) + q_b(t)}{2}##
##x_2 = \frac{q_p(t) - q_b(t)}{2}##

I'm stuck there. I still not sure about ##x_2## position and I can't see when the oscillations of the pendulum A will be at the minimum amplitude for the first time.

lmu6yjh-png.png
 
Last edited by a moderator:
Physics news on Phys.org
  • #4
EpselonZero said:
##q_p = 5 + ?, q_b = 5 - ?##
Why the signs and question marks? You have x1(0)=5 and x2(0)=0, no?
But you cannot assume the pure cosine formula for both. There may be a phase difference.
 
  • #5
haruspex said:
Why the signs and question marks? You have x1(0)=5 and x2(0)=0, no?
But you cannot assume the pure cosine formula for both. There may be a phase difference.
I'm not sure. because the statement say pendulum B at ##x_a = 0## so I guess it is not at the initial position. I think the initial position for the pendulum b is ##x_b = 0##
 
  • #6
EpselonZero said:
I'm not sure. because the statement say pendulum B at ##x_a = 0## so I guess it is not at the initial position. I think the initial position for the pendulum b is ##x_b = 0##
Well I am rather confused by this statement:
"At t = 0, pendulum A = xa=5cm and pendulum B = xa=0."
I assume you meant "pendulum B = xb=0." and that x1 is the same as xa, etc.
If that's not what xa means, please explain.
 
  • #7
I'm as confuse as you.
That's why the question marks.
However, it's really pendulum A at ##x_a = 5 cm## pendulum B at ##x_a = 0## So I suppose the pendulum B is at the initial position of pendulum A.

My guess is if the pendulum A moves from ##x_a = 0## to ##x_a = 5cm## and the pendulum B moves from ##x_b = 0 ### to ##x_a = 0## the pendulum b has also moved of 5cm. Finally it doesn't make sense. I don't have the distance between the pendulum A and B.Maybe pendulum B = pendulum A - 5cm

I assumed that A = ##x_a = 5## and B = ##x_b = 0##

I get

##x_a(t) = \frac{5}{2}(cos \omega_p t + cos \omega_b t)##
##x_b(t) = \frac{5}{2}(cos \omega_p t - cos \omega_b t)##

However, what should be the next step to find the first time the pendulum A reach the minimal amplitude.
 
Last edited by a moderator:
  • #8
EpselonZero said:
it doesn't make sense. I don't have the distance between the pendulum A and B.
I get

##x_a(t) = \frac{5}{2}(cos \omega_p t + cos \omega_b t)##
##x_b(t) = \frac{5}{2}(cos \omega_p t - cos \omega_b t)##

However, what should be the next step to find the first time the pendulum A reach the minimal amplitude.
I don't think the distance between their rest points matters, as long as they don't collide! We are only concerned with the displacement of each from its rest point.

The puzzle is, what is meant by reaching a minimum amplitude?

We could write each displacement in 'beat' form, i.e. as a product of two trig functions, and take the higher frequency factor as the oscillation and the (magnitude of the) lower frequency factor as representing a varying amplitude. The first minimum would then be where that magnitude is first zero.

Alternatively, we could consider the instantaneous amplitude as only meaningful at a local extremum of displacement, i.e. whenever the velocity is zero. These points will form a time sequence, and the displacement magnitudes at these points will vary. A minimum amplitude would then be a local minimum in that sequence.

These two interpretations will probably not lead to the same answer.

I would go with the beat interpretation and see where it leads.
 
  • #9
Do you mean something like
##x_a(t) = \frac{5}{2} cos (\frac{\omega_p + \omega_b}{2})t + cos (\frac{\omega_p - \omega_b}{2})t##

## cos (\frac{\omega_p - \omega_b}{2})t## is the lower frequency factor

## cos (\frac{\omega_p - \omega_b}{2})t = 0## when ##\omega_p = \pi + \omega_b##

If I plug those values I still don't know ##x_a(t)## and of course t.

I'm wondering why the first minimum would be where that magnitude is first zero.
 
  • #10
EpselonZero said:
Do you mean something like
##x_a(t) = \frac{5}{2} cos (\frac{\omega_p + \omega_b}{2})t + cos (\frac{\omega_p - \omega_b}{2})t##

## cos (\frac{\omega_p - \omega_b}{2})t## is the lower frequency factor
No, factor, not one term in a sum. And you left out several parentheses.
You have to write ##\frac{5}{2}( cos ((\frac{\omega_p + \omega_b}{2})t)+ cos ((\frac{\omega_p - \omega_b}{2})t))## as a product of two trig functions, not as a sum.
 
  • #11
I made a mistake
##x_a(t) = \frac{5}{2} cos (\frac{\omega_p + \omega_b}{2})t cos (\frac{\omega_p - \omega_b}{2})t##
 
  • #12
As a general approach we may make Lagrangean of the system
[tex]L=T-V[/tex]
[tex]T=\frac{1}{2}m_1l^2_1 \dot{\theta_1}^2+\frac{1}{2}m_2l^2_1 \dot{\theta_2}^2[/tex]
[tex]V=-g[m_1l_1 \cos\theta_1+m_2l_2 \cos\theta_2]+\frac{k}{2}d^2[/tex]
where
[tex]d=\sqrt{(l_1\cos\theta_1-l_2 \cos\theta_2)^2+(l_1\sin\theta_1-l_2 \sin\theta_2-R)^2}-l_0[/tex]
R is distance between the pendulum supports, ##l_0## is natural length of spring.
Applying Lagrange equation with some initial condition we get ##\theta_1(t),\theta_2(t)##.

I do not catch intention of the problem but it may be better to introduce
[tex]\phi_1=\frac{\theta_1+\theta_2}{2}[/tex]
[tex]\phi_2=\frac{\theta_1-\theta_2}{2}[/tex]
to get ##L(\phi_1,\phi_2,\dot{\phi_1},\dot{\phi_2})## for considering it.
 
  • #13
EpselonZero said:
I made a mistake
##x_a(t) = \frac{5}{2} cos (\frac{\omega_p + \omega_b}{2})t cos (\frac{\omega_p - \omega_b}{2})t##
Ok, but that makes it look as though the time variable is outside the trig function.
Clearer as ##x_a(t) = \frac{5}{2} \cos ((\frac{\omega_p + \omega_b}{2})t) \cos ((\frac{\omega_p - \omega_b}{2})t)##.
(In LaTeX you can put a backslash in front of standard functions, like cos, ln, to stop them being italicised.)
So which of those two factors do you consider to represent a varying amplitude, and when is it first zero?
 
  • #14
##\cos(\frac{\omega_p-\omega_b}{2}t)## represent a varying amplitude.

it is first zero when ##\frac{\omega_p-\omega_b}{2}t = \frac{\pi}{2}##

So when ##t = \frac{\pi}{\omega_p - \omega_b}##

Then I plug the value of t to find ##x_a(t)## ?
 
  • #15
EpselonZero said:
##\cos(\frac{\omega_p-\omega_b}{2}t)## represent a varying amplitude.

it is first zero when ##\frac{\omega_p-\omega_b}{2}t = \frac{\pi}{2}##

So when ##t = \frac{\pi}{\omega_p - \omega_b}##
Looks ok to me. Presumably they want a numeric answer, though.
EpselonZero said:
Then I plug the value of t to find xa(t) ?
Why? You know that will give 0. You are asked to find the time, which you have done.
 
  • #16
So that's it ? I mean, I'm not sure to understand why when this factor is zero, the amplitude of the oscillations is at minimum.
 
  • #17
EpselonZero said:
So that's it ? I mean, I'm not sure to understand why when this factor is zero, the amplitude of the oscillations is at minimum.
I would find it more convincing if the two modal frequencies were closer together, differing by no more than 15%, say. Then, plotting out the displacement of A would reveal a beat, and it would be reasonable to interpret the frequency difference as an amplitude modulation of an oscillation at the frequency sum.
But here your two frequencies are in a ratio 3:2. Still, that does put the sum and difference in the ratio 5:1, so I suppose it is just ok.
Plot it and see what it looks like.
 

Related to Finding the Oscillations of Pendulum A

1. What is a pendulum?

A pendulum is a weight suspended from a pivot point that can swing freely back and forth due to the force of gravity.

2. How do you find the oscillations of pendulum A?

To find the oscillations of pendulum A, you will need to measure the length of the pendulum, the angle of swing, and the time it takes for one full swing. Then, you can use the formula T = 2π√(L/g) to calculate the period of oscillation, where T is the time, L is the length of the pendulum, and g is the acceleration due to gravity.

3. Why is it important to find the oscillations of pendulum A?

Finding the oscillations of pendulum A can help us understand the behavior of pendulums and how they are affected by different factors such as length, angle, and gravity. This information can also be applied to other systems that exhibit oscillatory motion.

4. What factors can affect the oscillations of pendulum A?

The length of the pendulum, the angle of swing, and the force of gravity are the main factors that can affect the oscillations of pendulum A. Other factors such as air resistance and friction can also play a role.

5. How can the oscillations of pendulum A be used in real-life applications?

The oscillations of pendulum A have many practical applications, such as timekeeping in clocks and metronomes, and in seismology to measure the magnitude of earthquakes. They are also used in engineering to study the effects of vibrations on structures and in physics to demonstrate principles of harmonic motion.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
534
  • Introductory Physics Homework Help
Replies
9
Views
775
  • Introductory Physics Homework Help
Replies
27
Views
776
  • Introductory Physics Homework Help
Replies
4
Views
785
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
794
  • Introductory Physics Homework Help
Replies
26
Views
2K
Replies
1
Views
192
  • Introductory Physics Homework Help
Replies
14
Views
2K
Back
Top