Question

solve the attached question.

#JEE-NEET ​

Answer 1

AnswEr :

• The average kinetic energy during motion from the position of equilibrium topic the end is π²ma²v². [Option 2]

ExplanaTion :

We need to take the average kinetic energy to know the position of equilibrium topic the end.

$\sf : \implies \: \: \: KE_{avg} = \dfrac{KE_{max} +KE_{min} }{2}$

$\sf : \implies \: \: \: KE_{avg} = \dfrac{1}{2} KE_{max}$

$\sf : \implies \: \: \: KE_{avg} = \dfrac{1}{4} m { \omega}^{2} {a}^{2}$

$\sf : \implies \: \: \: KE_{avg} = \dfrac{1}{4} m { (2\pi v)}^{2} {a}^{2}$

$\sf : \implies \: \: \: KE_{avg} = \dfrac{1}{4} \times m \times 4 {\pi}^{2} \times {v}^{2} \times {a}^{2}$

$\sf : \implies \: \: \: { \purple{KE_{avg} = {\pi}^{2} m \: {a}^{2} \: {v}^{2}}}$

Therefore, the average kinetic energy during motion from the position of equilibrium topic the end is π²ma²v².

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More about the topic :

Formula's related to KE :

• KE = ½mv²

• KE = ½mω² (A²- x²)

• KE = ½mω² A²cos²ωt

• KE of the SHO is zero at the extreme position.

• Kinetic energy in SHM is also the periodic with double the frequency than that of displacement, velocity and acceleration .

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