Simple Pendulum?

Question

The length of a simple pendulum is 0.66m, the pendulum bob has a mass of 310 grams, and it is released at an angle of 12 degree to the vertical.

1. What is the pendulum bob's speed when it passes through the lowest point of the swing?

2. What is the total energy stored in this oscillation, assuming no losses?

Please show me how to do these problems step by step. Thank you very much. =)

Answer 1

The speed of the pendulum will be the maximum kinetic energy velocity. What you need to do is convert the potential energy to kinetic energy ie mgh=1/2*mv^2.....mass cancels from both sides... a little algebra and you get v= Square root(2gh). You get the height by the difference in the y-coordinates. Setup a triangle with a 12 degree angle to the vertical. Take .66m- .66m *cos(12)=.0144m
So v=Square root (2*9.80m/s^2*.0144m)=0.532 m/s

Total energy is equal to potential energy + kinetic energy...Since there is no kinetic energy, its just equal to potential energy...which is m*g*h=310 g*1kg/1000g*9.80m/s^2*.0144m=0.0437 J

Answer 2

in the straightforward pendulum, even as it vibrates there are 2 motions latest, one vibratory action in the bob different twisting action in the thread with which the bob is suspended.yet in case of compound pendulum, it includes a inflexible bar which vibrates backward and ahead alongside a level on it with which it truly is suspended. even as it vibrates, in reality a unmarried type of action i.e. vibratory action exists.

Answer 3

It's a change from potential to kinetic energy and back again.

P.E. = mgh
K.E. = 1/2 m v^2
h = 0.66 ( 1 - cos(12) )

1) P.E. = K.E. => gh = 1/2 v^2 => v = sqrt(2gh)
2) P.E. = mgh