What is the total energy stored in this oscillation

In summary, the length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12-degree to the verticle. With what frequency does it vibrate? The pendulum bob's speed when it passes through the lowest point of the swing is 10.5 cm/s. The total energy stored in this oscillation is 88.5 J.
  • #1
MIA6
233
0
1. The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12-degree to the verticle. (a) With what frequency does it vibrate? Assume SHM. b) What is the pendulum bob's speed when it passes through the lowest point of the swing? c) What is the total energy stored in this oscillation, assuming no losses?

For a), I used the formula f=1/2pai *(g/L)^(1/2) {I can't type the symbol 3.1415.. and radical} I wasn't sure because I didn't use the angle. For b), I used the formula v=radical F/(m/L). I don't think it is right, either, because of the angle. For c) E=1/2mv^2 + 1/2 kx^2, but how do I find x?
Thanks for help.
 
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  • #2
a) I agree

For parts b and c you really need to find the equation of motion. Since you probably are not familiar with differential equations then I will just tell you what it is.

[tex]x(t) = Acos(\omega t + \delta)[/tex]

Where A and [itex]\delta[/itex] are constants to be determined from your initial conditions, and omega (the w thing) is 2πf. It looks like you are supposed to assume that the pendulum is released from rest i.e. [itex]\frac{dx}{dt}[/itex](0) = 0, and that you are supposed to figure out the initial position x(0), where you would want to use that angle you are worried about.

I know that a lot people teach physics as, formula this, formula that, but try and start from the basics before resorting to formulas.
 
  • #3
Mindscrape said:
a) I agree

For parts b and c you really need to find the equation of motion. Since you probably are not familiar with differential equations then I will just tell you what it is.

[tex]x(t) = Acos(\omega t + \delta)[/tex]

Where A and [itex]\delta[/itex] are constants to be determined from your initial conditions, and omega (the w thing) is 2πf. It looks like you are supposed to assume that the pendulum is released from rest i.e. [itex]\frac{dx}{dt}[/itex](0) = 0, and that you are supposed to figure out the initial position x(0), where you would want to use that angle you are worried about.

I know that a lot people teach physics as, formula this, formula that, but try and start from the basics before resorting to formulas.

I know that formula, but first I have to find the amplitude? and it kinda relates to the angle? but you said when x(0)=12 degree which will convert to radians?
 
  • #4
For parts b and c, consider conservation of energy.
 
  • #5
Doc Al said:
For parts b and c, consider conservation of energy.

Is the speed at the bottom the greatest? But how do I know that at the bottom, it's the equilibrium point? And i don't know the amplitude, either.
 
  • #6
MIA6 said:
Is the speed at the bottom the greatest?
You tell me. Where is PE the lowest?
But how do I know that at the bottom, it's the equilibrium point?
Equilibrium has nothing to do with it.
And i don't know the amplitude, either.
What's the initial height of the bob, compared to the lowest point?
 
  • #7
Okay. PE is definitely the lowest at bottom because h=0, so the speed is at its maximum. But I don' know the initial height of the bob.
 
  • #8
MIA6 said:
PE is definitely the lowest at bottom because h=0, so the speed is at its maximum.
Good.
But I don' know the initial height of the bob.
Figure it out. You're given the string length and the angle for a reason! :wink:
 
  • #9
Can I use the formula: x=Lcos(angle)?
 
  • #10
MIA6 said:
Can I use the formula: x=Lcos(angle)?
That's almost what you need. That's not the height, but it will help you figure out the height. Draw yourself a diagram.
 
  • #11
ehh,, to find the height in triangle?
 
  • #12
Hint: The bottom position is a distance L below the pivot. How far below the pivot is the initial position? (The difference is the change in height.)
 

1. What is the definition of total energy stored in an oscillation?

The total energy stored in an oscillation refers to the sum of kinetic and potential energy within the system at a given point in time. It is a measure of the total energy contained in the oscillating system.

2. How is the total energy of an oscillation calculated?

The total energy of an oscillation can be calculated using the equation E = 1/2kA^2, where E is the total energy, k is the spring constant, and A is the amplitude of the oscillation.

3. What factors affect the total energy stored in an oscillation?

The total energy stored in an oscillation is affected by the mass of the object, the amplitude of the oscillation, and the spring constant of the system. Changes in these factors can alter the total energy stored.

4. Why is it important to know the total energy in an oscillation?

Knowing the total energy in an oscillation is important because it can give insight into the behavior and stability of the system. It can also help in determining the maximum displacement and velocity of the object in the oscillation.

5. Can the total energy in an oscillation ever be negative?

No, the total energy in an oscillation cannot be negative. This is because energy is a scalar quantity and cannot have a negative value. However, the total energy can decrease over time due to energy dissipation through friction or other forces.

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