Question

The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal,when displacement is ?

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Answer 1

Explanation:

$K.E.=P.E.⇒21mω2(a2−x2)=21mω2x2⇒a2−x2=x2⇒2x2=a2⇒x=a/2</p><p></p><p></p><p>$

Answer 2

Kinetic Energy = Potential Energy

KE = PE

$\frac{1}{2}m{ω}^{2}( {a}^{2} - {x}^{2} ) = \frac{1}{2} m{ω}^{2}{x}^{2}$

where,

'm' is the mass

'ω' is the velocity

'a' is the amplitude and

'x' is the displacement

$= ) \\ \\ {a}^{2} - {x}^{2} = {x}^{2} \\ \\ {a}^{2} = {x}^{2} + {x}^{2} \\ \\ {a }^{2} = 2 {x}^{2} \\ \\ {x}^{2} = \frac{ {a}^{2} }{2} \\ \\ x = \frac{a}{ \sqrt{2} }$

From this it is clear that,

The KE=PE in a SHM, when displacement is

a/√2