Question

A linear harmonic oscillator has a total mechanical energy of 200 J. Potential energy of it at mean position is 50J. Find (i) the maximum kinetic energy, (ii)the minimum potential energy, (iii) the potential energy at extreme positions.

Answer 1

The maximum kinetic energy = 150J

Minimum potential energy = 50J

The potential energy at extreme positions

= 200J

Given:

A linear harmonic oscillator has a total mechanical energy of 200 J.

Potential energy of it at mean position is 50J.

To find:

(i) the maximum kinetic energy.

(ii)the minimum potential energy.

(iii) the potential energy at extreme positions.

Concept used:

Total mechanical energy.

kinetic energy + potential energy = constant

Explanation:

At mean position, potential energy is minimum and kinetic energy is maximum. Hence,

$U_{min}$ = 50J (at mean position)

$K_{max}$= E−'$U_{min}$ =200−50

        = 150J            (at mean position)

At extreme positions kinetic energy is zero and potential energy is maximum

∴$U_{max}$ = E

         =200J         (at extreme position)

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