Question

In an inertia balance, a body supported against gravity executes simple harmonic oscillations in a horizontal plane under the action of a set of springs. If a 1.00 kg body vibrates at 1.00 Hz, a 2.00 kg body will vibrate at1) 0.500 Hz.2) 0.707 Hz.3) 1.00 Hz.4) 1.41 Hz.5) 2.00 Hz.

Answer 1

Answer:

2) f = 0.707 Hz

Explanation:

Given m₁ = 1.0 kg , f₁ = 1.0 Hz

So using the equation

f₁ = ( 1 / 2 π ) * √K / m₁

Solve to determine K' constant of spring

K = m * ( 4 π ² * f ² )

K = 1.0 kg * ( 4 π ² 1.0² Hz )

K = 39.4784176

So given 2.0 kg the frequency can be find using formula

f₂ = ( 1 / 2 π ) * √K / m₂

f₂ = ( 1 / 2 π ) * √39.4784176 / 2.0 kg

f₂ = 0.707 Hz