Question

a particle perform linear shm of amplitude 5 CM at what displacement from its mean position will the potential energy of the potential energy of the particle be 20% of its total energy

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Answer 1

Given:

A particle perform linear shm of amplitude 5 cm.

To find:

At what displacement from its mean position will the potential energy of the particle be 20% of its total energy ?

Calculation:

Let the displacement from mean position be x.

$\therefore \: PE = 20\%( TE)$

$\implies \: \dfrac{1}{2} m { \omega}^{2} {x}^{2} = 20\% \bigg( \dfrac{1}{2} m { \omega}^{2} {a}^{2} \bigg)$

$\implies \: {x}^{2} = 20\%( {a}^{2} )$

$\implies \: {x}^{2} = \dfrac{20}{100} ( {a}^{2} )$

$\implies \: {x}^{2} = \dfrac{ {a}^{2} }{5}$

$\implies \: x = \dfrac{a}{ \sqrt{5} }$

$\implies \: x = \dfrac{5}{ \sqrt{5} }$

$\implies \: x = \sqrt{5} \:cm$

So , the displacement from mean position is √5 cm.

Answer 2

Answer:

$\: x \ = \: \sqrt{5} \: cm$

Explanation:

PE=20%(TE)

$\frac{1}{2} mw {}^{2} x {}^{2} = 20\%( \frac{1}{2} mw {}^{2} a {}^{2} )$