Question

A mass of 5kg is suspended from the Free and of a spring when set for vertical oscillation the system execute 100 oscillation in 40 secons calculate the force constant of the spring

Answer 1

Answer:

We know that block has maximum speed at equilibrium position.

By applying conservation of mechanical energy at equilibrium position and highest point of oscillation:-

Answer 2

Given:

mass = 5kg

Number of oscillation = 100

time taken for oscillation = 40seconds

To Find:

The force constant of the spring

Explanation:

Energy at equilibrium-energy at highest point

1/2(mv²) +1/2kA²= mgA

The speed at highest point

ω = 2πf

  = 10πs⁻¹

ω = √k/m

   = k = mω²

At equilibrium-

mg =kA

A = mg/k = g/ω²

Now:-

1/2(mv²) + 1/2mω²A² = mgA

v² = 2gA - ω²A²

v² = (2g²/ ω²) - (g²/ω²)

    = g²/ω²

v² = 10²/(10π)² = 1/π²

v = 1/πm/s

Answer = 1/πm/s