ENERGY IN SIMPLE HARMONIC MOTION
For the simple harmonic motion, the force and the displacement are related by Hooke’s law
Since force is a vector quantity, in three dimensions it has three components. Further, the force in the above equation is a conservative force field; such a force can be derived from a scalar function which has only one component. In one dimensional case
As we have discussed in unit 4 of volume I, the work done by the conservative force field is independent of path. The potential energy U can be calculated from the following expression.
Comparing (10.63) and (10.64), we get
This work done by the force F during a small displacement dx stores as potential energy
From equation (10.22), we can substitute the value of force constant k = m ω2 in equation (10.65),
where ω is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation (10.6), we get
x = A sin ωt
Since the particle is executing simple harmonic motion, from equation (10.6)
x = A sin ωt
Therefore, velocity is
This variation with time is shown below.
Total energy is the sum of kinetic energy and potential energy
Alternatively, from equation (10.67) and equation
(10.72), we get the total energy as
E =1/2 mω 2 A 2 sin 2 ωt + 1/2 mω 2 A2 cos2 ωt
= 1/2 mω 2 A2 (sin2 ω t +cos2 ωt)
From trigonometry identity,
(sin2ωt + cos2ωt) = 1
E = 1/2 mω2 A2 = constant
which gives the law of conservation of total energy. This is depicted in Figure 10.26
Thus the amplitude of simple harmonic oscillator, can be expressed in terms of total energy.
Write down the kinetic energy and total energy expressions in terms of linear momentum, For one-dimensional case.
Kinetic energy is KE= 1/2 mvx2
Multiply numerator and denominator by m
KE= [1/2m] m2 vx2 = [1/2m] (mvx )2 = [1/2m] px2
where, px is the linear momentum of the particle executing simple harmonic motion.
Total energy can be written as sum of kinetic energy and potential energy, therefore, from equation (10.73) and also from equation (10.75), we get
E= KE +U( x) = [1/2m] px2 + 1/2 mω2 x2 = constant
Compute the position of an oscillating particle when its kinetic energy and potential energy are equal.
Since the kinetic energy and potential energy of the oscillating particle are equal,
1/2 mω 2 (A2 − x 2 ) = 1/2 mω2 x2
A2 − x2 = x2
2x2 = A2
x = ±A/√2