Question 157: Oscillations

Question

A particle of mass $$m$$ executes simple harmonic motion with amplitude a and frequency $$v.$$ The average kinetic energy during its motion from the position of equilibrium to the end is

Accepted Answer

KEY CONCEPT :
The instantaneous kinetic energy of a particle executing $$S.H.M.$$ is given by
$$K = {1 \over 2}m{a^2}{\omega ^2}{\sin ^2}\omega t$$
$$	herefore$$ average $$K.E. =    =   $$
$$ = {1 \over 2}m\omega {}^2{a^2}  $$
$$ = {1 \over 2}m{\omega ^2}{a^2}\left( {{1 \over 2}} ight)$$
$$\left( \, ight.$$ as $$\left. {   = {1 \over 2}} ight)$$
$$ = {1 \over 4}m{\omega ^2}{a^2} = {1 \over 4}m{a^2}{\left( {2\pi v} ight)^2}$$
$$\left( \, ight.$$ $$\left. {\omega  = 2\pi v} ight)$$
or,  $$\,\,\,\,\,   = {\pi ^2}m{a^2}{v^2}$$

Answer

$$2{\pi ^2}\,m{a^2}{v^2}$$

Answer

$${\pi ^2}\,m{a^2}{v^2}$$

Answer

$${1 \over 4}\,m{a^2}{v^2}$$

Answer

$$4{\pi ^2}m{a^2}{v^2}$$

Oscillations

A particle of mass $m executes simple harmonic motion with amplitude a and frequency v.$ The average kinetic energy during its motion from the position of equilibrium to the end is

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