Understanding Simple Gravity Pendulum - Urgent Help Needed

In summary: For question (b), I assume that the larger the angle A, the smaller the magnitude of F[v1]. Why is that? Think about it this way. The more angles A there are, the greater the number of vectors F[v1]. Each vector has a magnitude and direction, and the more vectors there are, the greater the magnitude and direction of each vector. So the more angles there are, the smaller the magnitude of F[v1] will be. For question (c), I assume that the acceleration vector a is not linear with the distance L. Why is that? Think about it this way. If the acceleration vector a was linear with distance L, then the o
  • #1
kajalove
15
0
hi

I know about harmonic oscilation, but I'm having trouble understanding how we derived formulas for gravity pendulum. Please read on.



If a ball on a string ( string is attached to the ceiling ) is displaced from its equilibrium position by angle A1, then forces on this ball are force of string F[v] and F[g].

F[v1] ... component of F[v] parallel to F[g] and of opposite direction to F[g]

http://img473.imageshack.us/img473/1854/nihaloje2.th.png

BTW - if picture doesn't show up then please look at the attached jpg file





1)

a)
Now why ( when angle A1 > 0 ) isn't the magnitude of F[v1] equal to F[g] --> F[v1] = -F[g]?


b)
I assume bigger the angle A, smaller is F[v1]. Why?
I assume it's because F[v] is constant no matter what the angle A is, but why is that?





2)
According to my book angles A and A1 are the same:

Code:
[B]F[net] = m * g * sin[A1]  =  m * g * sin[A][/B].

I'd imagine angle A being the same as angle A1 only if F[g] = F[v1]. Then direction of F[net] would be horizontal. But since that is not the case thus the two angles shouldn't be the same.





3)
I will quote my book:

Distance of a ball from equilibrium state can be stated with

Code:
[B]L = A1 * d = A * d [/B]
, where L is arc of a circle. When at angle A, the net force on the ball is F = m * g * sin[A], which gives the ball acceleration
Code:
[B]a = - g * sin[A][/B]

Acceleration vector a certainly isn't linear with L = A * d, and thus the osciliation isn't harmonic. But it becomes harmonic, if angle A is small enough for us to replace sin[A] with A




a)
a certainly isn't linear with L = A * d
I assume by that they mean to say that when arc L is twice as great, a isn't twice as great.
But what has that got to do with harmonic osiclation? Is with harmonic oscillation a linear with L?
Can you show me some proof of that?





b)
But it becomes harmonic, if angle A1 is small enough for us to replace sin[A1] with A1
First of all, I'm not sure that sin[A1] and A1 are ever roughly the same size, since no matter how small A1 is, sin[A1] will always be 100 or more times smaller. Right?




c)
Second, even if sin[A1] and A1 have about the same value when A1 is small enough, what is the purpose of replacing sin[A1] with A1? Why do we want to do that?





d)
Also, why is acceleration vector a negative?
I realize that when a has opposite direction to ball's velocity that it has to be negative. But sometimes ball's velocity and acceleration vectors have same direction and thus a should be positive?



cheers
 

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  • #2
Most of your questions can be answered by drawing a free body diagram. For question (a) consider what would happen if Fv1 = -Fg?
 
  • #3
Hootenanny said:
Most of your questions can be answered by drawing a free body diagram. For question (a) consider what would happen if Fv1 = -Fg?

Ball would start moving in a horizontal direction. I realize that and I know this is not the case, but why doesn't it happen?!
 
  • #4
Um if your angle keeps increasing it eventually reaches 90, so it keeps getting smaller and when it reaches 90 it goes to zero and tension is only determined in x direction since your tension component is basically Tension*cos(theta). I hope that answered the angle part of your problem
 

Related to Understanding Simple Gravity Pendulum - Urgent Help Needed

1. What is a simple gravity pendulum?

A simple gravity pendulum is a weight (called a bob) attached to a string or rod that is able to swing freely from a fixed point. It is often used as a model to understand the behavior of more complex pendulum systems.

2. How does a simple gravity pendulum work?

A simple gravity pendulum works by utilizing the force of gravity to create a back-and-forth swinging motion. As the bob is released from a certain height, it will swing back and forth in a predictable pattern until it comes to rest due to friction and air resistance.

3. What factors affect the period of a simple gravity pendulum?

The period (or time it takes for one swing) of a simple gravity pendulum is affected by three factors: the length of the string, the mass of the bob, and the strength of gravity. A longer string, larger mass, and stronger gravity will all result in a longer period.

4. How can I calculate the period of a simple gravity pendulum?

The period of a simple gravity pendulum can be calculated using the equation T = 2π√(L/g), where T is the period in seconds, L is the length of the string in meters, and g is the acceleration due to gravity (9.8 m/s² on Earth).

5. What are some real-world applications of the simple gravity pendulum?

The simple gravity pendulum has many real-world applications, including timekeeping devices (such as grandfather clocks), seismometers for measuring earthquakes, and amusement park rides. It is also used in physics classrooms as a teaching tool for understanding concepts such as potential and kinetic energy, periodic motion, and oscillations.

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