Question

Show average ke and average pe in one oscillation of a body having shm are equal

Answer 1

Answer:

To prove:

Average KE and Average PE in one Oscillation of a body having SHM are equal.

Proof:

Potential energy in SHM at any point is given as :

$\boxed{ \sf{ \red{PE = \frac{1}{2} m { \omega}^{2} {x}^{2}}}}$

So , average Potential energy is given as the mean of the max and min PE.

• Max PE is at amplitude
• Min PE is at Mean position

$\sf{ \red{PE_{avg} = \dfrac{\frac{1}{2} m { \omega}^{2} {A}^{2} + 0}{2}}}$

$\sf{ \red{PE_{avg} = \dfrac{\frac{1}{2} m { \omega}^{2} {A}^{2} }{2}}}$

$\sf{ \red{PE_{avg} = \dfrac{1}{4} m { \omega}^{2} {A}^{2}}}$

Kinetic energy is given as follows :

$\boxed{ \sf{ \red{KE = \frac{1}{2} m { \omega}^{2} ({A}^{2} - {x}^{2})}}}$

• Max KE at Mean position
• Min KE at Amplitude.

$\sf{ \red{KE_{avg} = \dfrac{0 + \frac{1}{2} m { \omega}^{2} {A}^{2} }{2}}}$

$\sf{ \red{KE_{avg} = \dfrac{ \frac{1}{2} m { \omega}^{2} {A}^{2} }{2}}}$

$\sf{ \red{KE_{avg} = \frac{1}{4} m { \omega}^{2} {A}^{2} }}$

So, we can say that :

$\boxed{ \bold{\sf{ \green{PE_{avg} = KE_{avg}}}}}$

Answer 2

$\blue{\underline{ \huge{ \blue{\boxed{ \mathfrak{\fcolorbox{red}{orange}{\purple{Answer}}}}}}}} \\ \\ \star \rm \: \red{To \: Prove}$

Average kinetic energy and Average potential energy of a body having SHM are equal...

$\star \rm \: \red{Formula}$

Formula of Kinetic energy of body havind SHM is given by...

$\dagger \: \boxed{ \rm{ \pink{ \bold{KE = \frac{1}{2} m{ { \omega}^{2} } ( {A}^{2} - {x}^{2} )}}}}$

Formula of potential energy of body having SHM is given by...

$\dagger \: \boxed{ \pink{ \rm{ \bold{PE = \frac{1}{2} m{ { \omega}^{2} } {x}^{2} }}}}$

$\star \rm \: \red{Calculation}$

• Max KE of body having SHM is at mean position and Min KE is at amplitude position..
• Max PE of body having SHM is at amplitude position and Min PE is at mean position...
• For amplitude position x = A and for mean position x = 0

$\leadsto \rm \: KE{ \tiny{av}} = \frac{KE{ \tiny{min}} + KE{ \tiny{max}}}{2} = \frac{0 + \frac{1}{2}m{ { \omega}^{2} } {A}^{2} }{2} \\ \\ \leadsto \: { \boxed{ \rm{ \bold{ \blue{KE{ \tiny{av}} = \frac{1}{4}m{ { \omega}^{2} } {A}^{2} }}}}} \\ \\ \leadsto \rm \: PE{ \tiny{av}} = \frac{PE{ \tiny{min}} + PE{ \tiny{max}}}{2} = \frac{0 + \frac{1}{2} m{ { \omega}^{2}} {A}^{2} }{2} \\ \\ \leadsto \: \boxed{ \blue{ \bold{ \rm{PE{ \tiny{av}} = \frac{1}{4} m{ { \omega}^{2} } {A}^{2} }}}} \\ \\ \dagger \: \boxed{ \bold{ \orange{ \rm{KE{ \tiny{av}} = PE{ \tiny{av}} = \frac{1}{4} m{ { \omega}^{2} } {A}^{2} }}}} \: \dagger$