An object of mass 0.2 kg executes simple har | vedclass

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(C) The total mechanical energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$ at any posit

Vedclass · Accepted Answer

$0.08$

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(C) The total mechanical energy $E$ of a particle in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$ at any position $x$.$E = K + U = 1 \ J + 0.6 \ J = 1.6 \ J$.The potential energy of a particle in $SHM$ is given by $U = \frac{1}{2} k x^2$,where $k$ is the force constant.We know that $k = m \omega^2$,where $\omega = 2 \pi f$.Given $f = \frac{25}{\pi} \ Hz$,so $\omega = 2 \pi (\frac{25}{\pi}) = 50 \ rad/s$.Thus,$k = 0.2 	imes (50)^2 = 0.2 	imes 2500 = 500 \ N/m$.Alternatively,we can use the total energy formula $E = \frac{1}{2} k A^2$,where $A$ is the amplitude.$1.6 = \frac{1}{2} 	imes 500 	imes A^2$.$1.6 = 250 	imes A^2$.$A^2 = \frac{1.6}{250} = \frac{16}{2500} = 0.0064$.$A = \sqrt{0.0064} = 0.08 \ m$.

An object of mass $0.2 \ kg$ executes simple harmonic oscillations along the $X$-axis with a frequency of $(\frac{25}{\pi}) \ Hz$. At the position $x=0.04 \ m$,the object has kinetic energy $1 \ J$ and potential energy $0.6 \ J$. The amplitude of oscillation is (in $m$)

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