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10 December, 12:57

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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  1. 10 December, 16:08
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    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
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