An object of mass 0.2 kg executes simple har | vedclass

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Question

(B) The total mechanical energy $E$ of a simple harmonic oscillator is the sum of its kinetic energy $K$ and potential energy $U$ at any position $x$.

Vedclass · Accepted Answer

$0.06$

Vedclass · Answer

(B) The total mechanical energy $E$ of a simple harmonic oscillator is the sum of its kinetic energy $K$ and potential energy $U$ at any position $x$.$E = K + U = 0.5 \ J + 0.4 \ J = 0.9 \ J$.The formula for the total energy of an oscillator is $E = \frac{1}{2} m \omega^2 A^2$,where $m$ is the mass,$\omega$ is the angular frequency,and $A$ is the amplitude.Given frequency $f = \frac{25}{\pi} \ Hz$,the angular frequency $\omega = 2 \pi f = 2 \pi \left( \frac{25}{\pi} ight) = 50 \ rad/s$.Substituting the values into the energy equation:$0.9 = \frac{1}{2} 	imes 0.2 	imes (50)^2 	imes A^2$$0.9 = 0.1 	imes 2500 	imes A^2$$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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