A block of mass 1 kg is fastened to a spring | vedclass

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(A) Given: Mass $m = 1 \ kg$,spring constant $k = 100 \ N \ m^{-1}$,amplitude $A = 10 \ cm = 0.1 \ m$,displacement $x = 5 \ cm = 0.05 \ m$.Angular fre

Vedclass · Accepted Answer

$0.375 \ J, 0.125 \ J$

Vedclass · Answer

(A) Given: Mass $m = 1 \ kg$,spring constant $k = 100 \ N \ m^{-1}$,amplitude $A = 10 \ cm = 0.1 \ m$,displacement $x = 5 \ cm = 0.05 \ m$.Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{1}} = 10 \ rad \ s^{-1}$.Velocity at displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2} = 10 \sqrt{0.1^2 - 0.05^2} = 10 \sqrt{0.01 - 0.0025} = 10 \sqrt{0.0075} = 10 	imes 0.0866 = 0.866 \ m \ s^{-1}$.Kinetic energy $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} 	imes 1 	imes (0.866)^2 = 0.5 	imes 0.75 = 0.375 \ J$.Potential energy $P.E. = \frac{1}{2} k x^2 = \frac{1}{2} 	imes 100 	imes (0.05)^2 = 50 	imes 0.0025 = 0.125 \ J$.

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