A particle is executing S.H.M. with an amplitude of $4$ cm. At what position from equilibrium position, the kinetic energy is equal to potential energy

1

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2 \sqrt{2}

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Solution

The correct option is D $2 \sqrt{2}$ cm
KE=PE
$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2}$
$A^{2} - x^{2} = x^{2}$
$x = \frac{A}{\sqrt{2}}$
$∴ x = \frac{4}{\sqrt{2}}$
$= 2 \sqrt{2} c m$

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A particle is executing S.H.M. with an amplitude of 4 cm. At what position from equilibrium position, the kinetic energy is equal to potential energy

The correct option is D 2√2 cmKE=PE12mω2(A2−x2)=12mω2x2A2−x2=x2x=A√2∴x=4√2=2√2cm

A particle is executing S.H.M. with an amplitude of $4$ cm. At what position from equilibrium position, the kinetic energy is equal to potential energy

The correct option is D $2 \sqrt{2}$ cm
KE=PE
$\frac{1}{2} m ω^{2} (A^{2} - x^{2}) = \frac{1}{2} m ω^{2} x^{2}$
$A^{2} - x^{2} = x^{2}$
$x = \frac{A}{\sqrt{2}}$
$∴ x = \frac{4}{\sqrt{2}}$
$= 2 \sqrt{2} c m$