**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.

### Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

**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.

### Ad blocker detected

Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

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:

The post How to work with potential energy in simple harmonic motion appeared first on YouPhysics