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.

Question

Bradley Odom

Physics

29 April, 23:49

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.

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Answers (1)

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Karissa Zuniga · Answer 1

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