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
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
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