Spring Energy Conservation

Classical Mechanics Level 1

A mass $m$ attached to a Hookean spring of spring constant $k$ is pulled so that the spring is displaced a distance $x$ from equilibrium. The mass is then released and allowed to oscillate at the end of the spring. What is the maximum velocity of the mass?

x^2\sqrt{k/m}

xk/m

kx^2/m

x\sqrt{k/m}

1 solution

Matt DeCross
Feb 2, 2016

By conservation of energy, the potential energy of the spring is entirely converted into kinetic energy of the mass when the mass is traveling at its maximum velocity. So the velocity $v$ obeys:

$\frac12 mv^2 = \frac12 kx^2.$

Solving for $v$ obtains the answer $v = x\sqrt{k/m}$ .

So many variables could apply, but where would be the benifit?

nadine perea - 3 years, 10 months ago

Spring Energy Conservation

1 solution

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