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The 2.50 kg cube in the figure has edge lengths d = 6.50 cm and is mounted on an axle
through its center. A spring (k = 900 N/m) connects the cube's upper corner to a rigid wall.
Initially the spring is at its rest length. If the cube is rotated 2.00° and released, what is the
period of the resulting SHM?​

The 250 kg cube in the figure has edge lengths d 650 cm and is mounted on an axlethrough its center A spring k 900 Nm connects the cubes upper corner to a rigid class=

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MathPhys

Answer:

0.191 s

Explanation:

The distance from the center of the cube to the upper corner is r = d/√2.

When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ.  The new vertical distance from the center to the corner is r cos θ.

Sum of the torques:

∑τ = Iα

Fr cos θ = Iα

(k r sin θ) r cos θ = Iα

kr² sin θ cos θ = Iα

k (d²/2) sin θ cos θ = Iα

For a cube rotating about its center, I = ⅙ md².

k (d²/2) sin θ cos θ = ⅙ md² α

3k sin θ cos θ = mα

3/2 k sin(2θ) = mα

For small values of θ, sin θ ≈ θ.

3/2 k (2θ) = mα

α = (3k/m) θ

d²θ/dt² = (3k/m) θ

For this differential equation, the coefficient is the square of the angular frequency, ω².

ω² = 3k/m

ω = √(3k/m)

The period is:

T = 2π / ω

T = 2π √(m/(3k))

Given m = 2.50 kg and k = 900 N/m:

T = 2π √(2.50 kg / (3 × 900 N/m))

T = 0.191 s

The period is 0.191 seconds.

batolisis

The period of the resulting SHM is : 0.191 sec

Given data :

Edge length of cube = 6.50 cm

Spring ( K )  = 900 N/m  

Radius ( r ) = [tex]\frac{d}{\sqrt{2} }[/tex]  

mass ( m ) = 2.50 kg

Determine the period of the resulting SHM

Distance stretched by the spring when the cube is stretched through θ =  r sin θ

Vertical distance from the center = r cos θ

F = k r sin θ

next step : determine the sum of torques

sum of torques ( ∑τ ) = I∝

                          F * r cos θ = I∝

                          ( k r sin θ ) r cos θ =  I∝  

                          k (d²/2) sin θ cos θ = I∝  ------ ( 1 )

Note : For cube rotating around its center I = 1/6 md²

Back to equation ( 1 )

k (d²/2) sin θ cos θ = 1/6 md² * ∝

[tex]\frac{3}{2}[/tex] k sin(2θ) = m∝     ( sin θ ≈ θ for small values like 2° )

∴ ∝ = ( 3k / m ) θ

d²θ/dt² = ( 3k / m ) θ

∴ Angular frequency ( w ) = √ (3k/m)

Final step : Calculate the value of the period of the resulting SHM

T = 2π / ω

  = 2π / √( m / (3k) )

  = 2π / √(2.50 kg / (3 * 900 N/m))

  = 0.191 sec

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