Question

Show that the total energy of particles performing Linear S.H.M is constant.

Answer 1

For these types of questions use the equations of S.H.M, $x = {\text{ A}}\sin \omega t$ and $v = {\text{A}}\omega \cos \omega t$ . Now use the equation of Conservation Of energy formulas and find the value of the total energy.We know the Kinetic Energy and Potential energy formula; we just have to find total energy. We need to know the standard definition of potential and kinetic energy.

Complete step by step answer:
We know that the formula for Kinetic Energy = $\dfrac{1}{2}m{v^2}$
Where m=mass and v= velocity of body. We have value of $v = {\text{A}}\omega \cos \omega t$
A= Amplitude of the given SHM, $\omega$ = given angular velocity. $t$ = given time.
Also, we know that the formula for Potential Energy= $\dfrac{1}{2}k{x^2}$
Where $k = {\omega ^2}m$ and $x = {\text{ A}}\sin \omega t$

Now as velocity is the rate of change of displacement, thus
$v = \dfrac{{dx}}{{dt}}$
$\Rightarrow v = {\text{ A}}\omega {\text{ }}cos{\text{ }}\omega t$
Thus, Total Energy = Kinetic Energy + Potential Energy = $\dfrac{1}{2}m{v^2} + {\text{ }}\dfrac{1}{2}k{x^2}$
Taking the value of the velocity and displacement,
$\Rightarrow$T.E = $\dfrac{1}{2}m{({\text{A}}\omega \cos \omega t)^2} + {\text{ }}\dfrac{1}{2}k{({\text{A}}\sin \omega t)^2}$
Substituting the value of k $= {\omega ^2}m$
$\Rightarrow T.E. = \dfrac{1}{2}m{({\text{A}}\omega \cos \omega t)^2} + {\text{ }}\dfrac{1}{2}{\omega ^2}m{({\text{A}}\sin \omega t)^2}$

Taking the common terms out of parenthesis, we get
$\Rightarrow T.E. = \dfrac{1}{2}{\omega ^2}m{{\text{A}}^2}{(\cos \omega t)^2} + {\text{ }}\dfrac{1}{2}{\omega ^2}m{{\text{A}}^2}{{\text{(}}\sin \omega t)^2}$
$\Rightarrow T.E. = \dfrac{1}{2}{\omega ^2}m{{\text{A}}^2}[{(\cos^2 \omega t)}+{(\sin^2 \omega t)}]$
Using the trigonometric identity, we get the final answer
$\Rightarrow T.E = \dfrac{1}{2}{\omega ^2}m{{\text{A}}^2}$.
Hence, we see that the total energy of the S.H.M system is a constant and is equal to $\dfrac{1}{2}{\omega ^2}m{{\text{A}}^2}$.
Here we see that this can be compared to an Energy system,$E=\dfrac{1}{2}m{v_0}^2$
Thus, we can easily say ${v_0}$= {\text{A}}$$$$\omega . Thus, we get the velocity for a given SHM. Hence proved.

Note: These types of questions demand the formula application of SHM. SHM has numerous formulas and each has to be remembered thoroughly, and also should be practiced with proof. We know that each formula in SHM is interconnected, thus it would be best if you can proof all the formulas from the given displacement equation, as it is always better to practice and remember questions than memorising physics equations. The base equation of displacement and velocity should always be remembered. The equation of total energy is also very important and should be remembered.