The Period of the resulting shm will be T=39.7
Explanation:
Given data
m=3kg
d=.06m
k=1200 N/m
Θ=3 °
T=?
we have the formulas,
I = (1/6)Md2
F = ma
F = -kx = -(mω2x)
k = mω2 τ = -d(FgsinΘ)
T=2 x 3.14/ √(m/k)
Solution for the given problem would be,
F=-Kx (where x= dsin Θ)
F=-k dsin Θ
F=-(1200)(.06)sin(3 °)
F=-10.16N
By newton's second law.
F = ma
a= F/m
a=(-10.16N)/3
a=3.38
using the k=mω value
k=mω
ω=k/m
ω=1200/3
ω=400
Using F = -kx value
x = F/-k
x=(-10.16)/1200
x=0.00847m
Restoring the torque value
τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =
α =(.06)(4)(9.81)sin(4°)
α=-1.781
Rotational to linear form
a = αr
r = .1131 m
a=-1.781 x .1131 m
a=-0.2015233664
Time Period
T=2 x 3.14/ √(m/k)
T=6.28/√(3/1200)
T=6.28/0.158
T=39.7
