**Problem statement:**

A mass M = 3 kg attached to a spring of constant (stiffness) k = 1000 N/m is lying on a frictionless surface. The spring is initially stretched, the mass is at rest and its total energy is E = 20J. At time t = 0 the mass is released. Find:

- The amplitude, angular frequency, frequency and period of the simple harmonic motion undergone by M.
- The acceleration of the mass at t = 0.
- The position and velocity of the mass when its potential energy 10 J.

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**Solution:**

At t = 0 the x-coordinate of the position of the mass is x = A, equal to the amplitude of the simple harmonic motion. Since it is at rest it has no kinetic energy and therefore its total energy equals its elastic potential energy:

From the previous equation we can isolate the amplitude A of the simple harmonic motion. Isolating and substituting the numerical values we get:

On the other hand, the angular frequency ω of the simple harmonic motion related to the constant k of the spring is given by:

And the relationship between the angular frequency ω, the frequency ν and the period T of the simple harmonic motion is:

Substituting the givens of the problem:

In the figure below is shown mass M at t = 0.

The force acting upon it is given by Hooke’s law and is proportional to the displacement *x* of the mass.

At t = 0 the displacement of the mass with respect to its equilibrium position x = 0 is the amplitude A. Therefore the force will reach its maximum at this point.

Moreover, Newton’s second law must hold, and therefore we can isolate the acceleration of the mass M:

And substituting the numerical values:

The acceleration of the mass can also be found by taking the second derivative of the position function with respect to time:

And substituting x = A in the previous formula:

As you can see, both methods yield identical results (in absolute values). The minus sign means that the acceleration vector points to the negative direction of the x-axis, and that is so because a restoring force tends to bring the mass to its equilibrium position x = 0.

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From the elastic potential energy (10 J) we can find the displacement x of the mass:

Its velocity can be found using the principle of conservation of energy . The initial energy of the mass equals its total energy:

And substituting the previously calculated value of x, as well as the amplitude of the simple harmonic motion we get:

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