Answer:
a. 0.9370 s =Ta
b. 0.8510 s = Tb
c . Error = 0.086 s = Ta -Tb
Step-by-step explanation:
a) This is a simple harmonic motion problem = T =2π√l/g
T = Period in s
I = length or distance l = 21.8 cm = 21.8/100 = 0.218 m note 100 cm = 1m
g = acceleration due to gravity = 9.81 m/s^ 2
π = pie = 22/7
substitute
T = 2* 22/7 *√0.218/9.81 = 0.937018962 s
approximate = 0.9370 s = Ta
b) As a point mass implies the entire object as it is concentrated in space neglecting its dimension in space
for a pendulum 2πf = [(MgL)/l]
l = moment of inertia of a solid sphere = 2/5MR^2
M = mass
R = radius = 14.0 cm = 14.0 / 100 = 0.14 m
f = frequency = 1/T
g = acceleration due to gravity = 9.81m/s^2
L = I = length or distance l = 21.8 cm = 21.8/100 = 0.218 m note 100 cm = 1m
Thus 2π1/T = [(MgL)/l ] ^1/2
make subject formula and substitute l = moment of inertia of a solid sphere = 2/5MR^2 in the equation
square both sides
4*π^2/ T^2 = [(MgL)/l ]
T^2 = 4*π^2 *l / [(MgL]
= 4*π^2* 2/5MR^2 /[(MgL]
= 8* π^2*R^2/gL
= 8* 22/7^2 *0.14^2/ 9.81* 0.218
= 0.7242188742
T =√0.7242188742 = 0.8510105018 s
approximate = 0.8510 s = Tb
Error = Ta - Tb = 0.9370 s - 0.8510 s = 0.086 s