Most as long the hypothesis is a good answer and can be answered
No, and no. In fact, the consequences are exactly opposite to your description.
When you drop soap on the ground, the soap ... which had been clean ... gets dirty, and the ground ... which had been dirty ... gets clean.
The Period of the resulting shm will be T=39.7
<u>Explanation:</u>
<u>Given data</u>
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
<u>By newton's second law.</u>
F = ma
a= F/m
a=(-10.16N)/3
a=3.38
<u>using the k=mω value</u>
k=mω
ω=k/m
ω=1200/3
ω=400
<u>Using F = -kx value</u>
x = F/-k
x=(-10.16)/1200
x=0.00847m
<u>Restoring the torque value </u>
τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =
α =(.06)(4)(9.81)sin(4°)
α=-1.781
<u>Rotational to linear form</u>
a = αr
r = .1131 m
a=-1.781 x .1131 m
a=-0.2015233664
<u>Time Period</u>
T=2 x 3.14/ √(m/k)
T=6.28/√(3/1200)
T=6.28/0.158
T=39.7