An object swings from the end of a cord as a simple pendulum with period T. An identical object oscillates up and down on the end of a vertical spring with the same period T. If the masses of both objects are doubled, how will the new values of the periods compare to T?

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An object swings from the end of a cord as a simple pendulum with period T. An identical object oscillates up and down on the end of a vertical spring with the same period T. If the masses of both objects are doubled, how will the new values of the periods compare to T?

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Answer 1 · 2014-11-28T17:33:14

The period of a pendulum can be calculated by knowing the acceleration of gravity and the length of the pendulum.

$T = 2 π \sqrt{\frac{L}{g}}$ $T = 2pi sqrt(L/g)$

The mass of the pendulum doesn't figure into the equation. While it is true that twice as much force will be required to accelerate a mass that is twice as large, the gravitational force increases as we increase the mass.

The period of a simple spring depends on the spring constant $k$ $k$ and the mass $m$ $m$ .

$T = 2 π \sqrt{\frac{m}{k}}$ $T = 2pi sqrt(m/k)$

Where m is the mass of the object and k is the spring constant. If we replace m with 2m, we can calculate that the period will increase by $\sqrt{2}$ $sqrt(2)$ .

$T' = \sqrt{2} T$ $T' = sqrt(2)T$ for the spring-mass system.

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An object swings from the end of a cord as a simple pendulum with period T. An identical object oscillates up and down on the end of a vertical spring with the same period T. If the masses of both objects are doubled, how will the new values of the periods compare to T?

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