Question

An 81.9 g mass is attached to a horizontal spring with a spring constant of 3.5 N/m and released from rest with an amplitude of 39.1 cm. What is the speed of the mass when it is halfway to the equilibrium position if the surface is frictionless? Answer in units of m/s.

Answer 1

The potential energy of the spring if it is pulled through x distance is 1/2 k x^2.
When it is released with this energy it gains speed.
Half the amplitude is x/2.
At half the amplitude its p.e is 1/2 k [x/2]^2 = 1/8 kx^2.
Hence the remaining energy is converted into kinetic energy of the mass m as 1/2 mv^2.

The remaining energy is 3/8 kx^2 = 1/2 mv^2
Substitute the values and find v.

3/8 *3.5* 0.391 ^2 = 1/2 *0.0819 v^2

v = - 2.21 m/s minus sign since it moves toward the equilibrium position.
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