A particle of mass 10, g moves in a field whe | vedclass

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Question

(A) The potential energy per unit mass is given by $v = \frac{1}{2} \omega^2 x^2$.Comparing this with the given expression $v = 8 	imes 10^4 x^2$,we

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$x = 10\, \sin(400\, t + \phi)\, 	ext{cm}$

Vedclass · Answer

(A) The potential energy per unit mass is given by $v = \frac{1}{2} \omega^2 x^2$.Comparing this with the given expression $v = 8 	imes 10^4 x^2$,we get $\frac{1}{2} \omega^2 = 8 	imes 10^4$.Thus,$\omega^2 = 16 	imes 10^4$,which gives $\omega = 400\, 	ext{rad/s}$.The total energy $E$ of the particle is given by $E = \frac{1}{2} m \omega^2 A^2$.Given $m = 10\, 	ext{g}$,$E = 8 	imes 10^7\, 	ext{erg}$,and $\omega = 400\, 	ext{rad/s}$.Substituting these values: $8 	imes 10^7 = \frac{1}{2} 	imes 10 	imes (400)^2 	imes A^2$.$8 	imes 10^7 = 5 	imes 160000 	imes A^2 = 800000 	imes A^2$.$A^2 = \frac{8 	imes 10^7}{8 	imes 10^5} = 100$.So,$A = 10\, 	ext{cm}$.The displacement equation is $x = A \sin(\omega t + \phi) = 10 \sin(400 t + \phi)\, 	ext{cm}$.

$A$ particle of mass $10\, g$ moves in a field where potential energy per unit mass is given by the expression $v = 8 \times 10^4\, x^2\, \text{erg/g}$. If the total energy of the particle is $8 \times 10^7\, \text{erg}$,then the relation between $x$ and time $t$ is:

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