Physics

E = $2mw_{2}(A_{2}−A_{2}sin_{2}(wt+ϕ)) $

$P.E.=2mw_{2}x_{2}sin_{2}(wt+ϕ) $

Total energy = $2mw_{2}(A_{2}) $ (Independent of time)

Solution:

$U_{0}=$minimum potential energy at mean position $(x=2)=10J$

At extreme position

$U=$ Total mechanical energy $=26J$

$=10+(x−2)_{2}$

$∴$ $(x−2)=±4$

Hence $x=6m$ and $x=−2m$ are the extreme positions.

Amplitude of oscillation$=4m$

Potential Energy is : $2mw_{2}(x_{2}) $

Total Energy is: $2mw_{2}(A_{2}) $

Example: The potential energy of a simple pendulum in its resting position is $10$ J and its mean kinetic energy is $5$ J. What will be its total energy at any instant?

Solution:

The total energy of the system remains constant. Since it is given that P.E at rest is $10$ J, the total energy must be $10$ J as K.E at rest is 0. As total energy of the system is conserved in SHM.

Solution:

At the extreme position of the spring it has only potential energy since velocity is zero: $2kA_{2} $

At the equilibrium position it has no stretch in the spring.

Kinetic energy at this instant: $2mu_{2} $

At any instant of time during the motion:

Total energy = KE + PE = $2mu_{2} $ = $2kA_{2} $

ExampleDefinitionsFormulaes

ExampleDefinitionsFormulaes