Question

A simple pendulum of length L = 10.0 m is released from an angle of .400rad. Assume the pendulum swings with simple harmonic motion.

c) Using conservation of energy(gravitational and kinetic), find the linear speed of the bob at its lowest point.

d) If the linear speeds found in (b) and (c) were exactly the same, explain why. If they were different explain why.

Answer 1

a)
Time period, T = 2*pi*sqrt(L/g)

= 2*pi*sqrt(10/9.8)

= 6.34 s

angular frequency, w = 2*pi/T

= 2*pi/T

= 2*pi/(6.34)

= 0.99 rad/s <<<<<<<<--------Answer

b) Dont know how to do

c) given,

theta = 0.4 rad

= 0.4*(360/2*pi)

= 23 degrees

now initial height of the bob above its equilibrium point,

h = L*(1 - cos(23))

= 10*(1 - cos(23))

= 0.795 m

As the boob comes to equilibrium point its potential energy is converted to kinetic energy

m*g*h = 0.5*m*v^2

==> v = sqrt(2*g*h)

= sqrt(2*9.8*0.795)

= 3.95 m/s <<<<<<<<<<<<-----------Answer

d) part a question is not given