The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con- nects the cube’s upper corner to a rigid wall. initially the spring is at its rest length. if the cube is rotated 3 and released, what is the period of the resulting shm?

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

area of focus