A simple pendulum of mass 200 grams has an angular amplitude

A simple pendulum of mass= 200 grams has an angular amplitude of 3 degrees and a period of 2 s. A.) Find the maximum tension in the string. B.) Write the displacement equation for this pendulum if at t=0 the pendulum was 2 degrees from vertical ( show the numerical values for amplitude, angular frequency and phase constant)

Solution

A)
   T = 2pi sqrt(L/g)

2 = 2pi sqrt(L / 9.8)

L = 1 m

maximum tension will be at bottom position.

for that first we have to find speed at bottom position.

height of poisition of mass at 3 deg .

h = L ( 1- cos@) = 1 ( 1 - cos3) = 1.37 x 10^-3 m

Using energy conservation to find speed at bottom:

m v^2 /2 = mgh

v = sqrt(2gh) = sqrt(2 x 9.8 x 1.37 x 10^-3) = 0.164 m/s

At bottom using Fnet = ma

T - mg = m (v^2/r )

T = 0.200x9.8 + 0.200(0.164^2 /1) = 1.97 N

b) theta = theta_max cos(wt + phi)

theta_max = 3 deg ,

w = 2pi/T = 2pi/2 = pi rad/s

theta = 3 deg cos(pi t + phi )

at t= 0

2 =3cos(phi)

phi = 48.19 deg OR 0.84 rad

theta = (3 deg ) cos(pi t + 0.84)