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A mass of 0. 40 kg, hanging from a spring with a spring constant of 160 n/m, is set into an up-and-down simple harmonic motion. What is the speed of the mass when moving through the equilibrium point? the starting displacement from equilibrium is 0. 10 m.

1 Answer

4 votes

Given:


m=0.40kg


k=160(N)/(m)


A=0.10m

We know that the velocity of a mass attached to a spring is greatest when passing through equilibrium. Using the following formulas we can compute velocity.


v_(max) =\pm Aw and
w=\sqrt{(k)/(m) }
\Longrightarrow v_(max) =\pm A\sqrt{(k)/(m) }


\Longrightarrow v_(max) =\pm A\sqrt{(k)/(m) }


\Longrightarrow v_(max) =\pm (0.10)\sqrt{(160)/(0.40) }


\Longrightarrow v_(max) =\pm (0.10)(20) }


\Longrightarrow v_(max) =\pm 2.0(m)/(s)

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