The total mechanical energy of a particle exe | vedclass

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(A) The total mechanical energy of a particle in simple harmonic motion is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is

Vedclass · Accepted Answer

$\frac{3}{4}E$

Vedclass · Answer

(A) The total mechanical energy of a particle in simple harmonic motion is given by $E = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.The kinetic energy $(K.E.)$ at any displacement $x$ is given by $K.E. = \frac{1}{2} k (A^2 - x^2)$.Given that the displacement $x = \frac{A}{2}$,we substitute this into the kinetic energy formula:$K.E. = \frac{1}{2} k \left( A^2 - (\frac{A}{2})^2 \right)$$K.E. = \frac{1}{2} k \left( A^2 - \frac{A^2}{4} \right)$$K.E. = \frac{1}{2} k \left( \frac{3A^2}{4} \right)$$K.E. = \frac{3}{4} \left( \frac{1}{2} k A^2 \right)$Since $E = \frac{1}{2} k A^2$,we have $K.E. = \frac{3}{4} E$.

The total mechanical energy of a particle executing simple harmonic motion is $E$. When the displacement is half the amplitude,its kinetic energy will be

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