Question

A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don’t want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25cm? In this case, how much is the energy per unit mass of the block? (g≈π2≈10m s-2)

Answer 1

ω²a <= g

Explanation:

Given: Mass of wooden block is m.

Amplitude of vibrations is restricted to 25cm

g = 10 m/s²

Find: Maximum frequency possible such that the block fully stays in contact with the piston.

Solution:

For the block to not separate from piston, spring force should not go greater than weight.

Energy per unit mass of the block.

Solution: ω is the angular frequency

α (max) = ω²a where ω is the angular frequency and a is acceleration.

α (max) <= g

meaning ω²a <= g

So acceleration should be less than or equal to 10 m/s²

We know that ω = k m where k is force constant. So ω is proportional to mass.

Answer 2

Answer:N= 1/s

E/m= 1.25 J/kg

Explanation:

For the block to not leave contact with the piston,

mg = kx

and (X)max = A

So, mg = kA

m/K = A/g

Now ,

T = 2π ×√(m/k)

= 2π ×√(A/g)

= 2π×√(25/100÷ π²)

= 2π ×√(1/4π²)

= 2π ×1/2π

= 1 s

So, n = 1/T

= 1 /s

Now,

E = 1/2 × mw²A²

E/m = 1/2 × w²A²

= 1/2 × (2π/1)² × (25/100)²

= 1/2 × 4π² × 1/16

= 1/2 × 4×10 × 1/16

= 20/16

= 5/4

= 1.25 J/kg