An object of mass 0.2 , kg executes simple ha | vedclass

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(B) The total mechanical energy $E$ of an object in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$.$E = K + U =

Vedclass · Accepted Answer

$0.06$

Vedclass · Answer

(B) The total mechanical energy $E$ of an object in simple harmonic motion is the sum of its kinetic energy $K$ and potential energy $U$.$E = K + U = 0.5 \, J + 0.4 \, J = 0.9 \, J$.The formula for the total energy of an oscillator is $E = \frac{1}{2} k A^2$,where $k = m \omega^2$.Given mass $m = 0.2 \, kg$ and frequency $f = \frac{25}{\pi} \, Hz$.The angular frequency $\omega = 2 \pi f = 2 \pi \left( \frac{25}{\pi} ight) = 50 \, rad/s$.The force constant $k = m \omega^2 = 0.2 	imes (50)^2 = 0.2 	imes 2500 = 500 \, N/m$.Using $E = \frac{1}{2} k A^2$,we have $0.9 = \frac{1}{2} 	imes 500 	imes A^2$.$0.9 = 250 	imes A^2$.$A^2 = \frac{0.9}{250} = 0.0036$.$A = \sqrt{0.0036} = 0.06 \, m$.

An object of mass $0.2 \, kg$ executes simple harmonic motion along the $X$-axis with a frequency of $\frac{25}{\pi} \, Hz$. At the position $x = 0.04 \, m$,the object has a kinetic energy of $0.5 \, J$ and a potential energy of $0.4 \, J$. The amplitude of oscillation in meters is equal to:

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