Question

Show that the total energy of the particle is SHM is constant at any point on it's path​

Answer 1

Answer

We know that,

$\sf Potential \: energy = \dfrac{1}{2} m \omega {}^{2}{y}^{2}$

$\sf Kinetic \: energy = \dfrac{1}{2} m \omega {}^{2} ({A}^{2} - {y}^{2})$

Case 1 :-

When the body is at displacement y

The total energy is given by,

$\sf \dashrightarrow TE = PE + KE$

$\sf \dashrightarrow TE = \dfrac{1}{2} m \omega {}^{2} {y}^{2} + \dfrac{1}{2} m \omega {}^{2} ({a}^{2} - {y}^{2})$

$\sf \dashrightarrow TE = \dfrac{1}{2} m \omega {}^{2} \bigg( {y}^{2} + ( {A}^{2} - {y}^{2}) \bigg)$

$\sf \dashrightarrow \boxed{\sf TE = \dfrac{1}{2} m \omega {}^{2} {A}^{2}} \: --- (i)$

Case 2 :-

When the body is at mean position y = 0

The total energy is given by,

$\sf \dashrightarrow TE = PE + KE$

$\sf PE = \dfrac{1}{2} m \omega {}^{2} ({y}^{2})$

$\sf PE = \dfrac{1}{2} m \omega {}^{2} (0)$

$\sf PE = 0$

$\sf KE = \dfrac{1}{2} m \omega {}^{2} ({A}^{2} - {y}^{2})$

$\sf KE = \dfrac{1}{2} m \omega {}^{2} ({A}^{2} - 0)$

$\sf KE = \dfrac{1}{2} m \omega {}^{2} {A}^{2}$

By substituting the values of KE and PE,

$\sf \dashrightarrow TE = 0 + \dfrac{1}{2} m \omega {}^{2} {A}^{2}$

$\sf \dashrightarrow \boxed{\sf TE = \dfrac{1}{2} m \omega {}^{2} {A}^{2}} \: --- (ii)$

Case 3. -

When the displacement is equal to 'A'

The total energy is given by,

$\sf TE = PE + KE$

$\sf PE = \dfrac{1}{2} m \omega {}^{2} {y}^{2}$

$\sf PE = \dfrac{1}{2} m \omega {}^{2} {A}^{2}$

$\sf KE = \dfrac{1}{2} m \omega {}^{2} ({A}^{2} - {y}^{2})$

$\sf KE = \dfrac{1}{2} m \omega {}^{2} ({A}^{2} - {A}^{2})$

$\sf KE = \dfrac{1}{2} m \omega {}^{2} (B)$

$\sf KE = 0$

By substituting the values of KE and PE,

$\sf \dashrightarrow TE = \dfrac{1}{2} m \omega {}^{2} {A}^{2} + 0$

$\sf \dashrightarrow \boxed{\sf TE = \dfrac{1}{2} \omega {}^{2} {A}^{2}} \: --- (iii)$

From equation (i), (ii) and (iii) total energy is, $\sf \dfrac{1}{2} m \omega {}^{2} {A}^{2}$ = constant

Thus the total energy of an object under SHM is constant at any point on it's path.