A spring executes S.H.M. with mass 10 text{ k | vedclass

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(B) Given:Mass $m = 10 	ext{ kg}$Force constant $k = 10 	ext{ N/m}$Velocity $v = 40 	ext{ cm/s} = 0.4 	ext{ m/s}$Amplitude $A = 0.5 	ext{ m}$Angu

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$0.3$

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(B) Given:Mass $m = 10 	ext{ kg}$Force constant $k = 10 	ext{ N/m}$Velocity $v = 40 	ext{ cm/s} = 0.4 	ext{ m/s}$Amplitude $A = 0.5 	ext{ m}$Angular frequency $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{10}{10}} = 1 	ext{ rad/s}$.The velocity of a particle in $S$.$H$.$M$. is given by $v = \omega \sqrt{A^2 - x^2}$, where $x$ is the displacement.Substituting the values: $0.4 = 1 	imes \sqrt{(0.5)^2 - x^2}$.Squaring both sides: $(0.4)^2 = (0.5)^2 - x^2$.$0.16 = 0.25 - x^2$.$x^2 = 0.25 - 0.16 = 0.09$.$x = \sqrt{0.09} = 0.3 	ext{ m}$.Thus, the displacement is $0.3 	ext{ m}$.

$A$ spring executes $S$.$H$.$M$. with mass $10 \text{ kg}$ attached to it. The force constant of spring is $10 \text{ N/m}$. If at any instant its velocity is $40 \text{ cm/s}$, the displacement at that instant is (Amplitude of $S$.$H$.$M$. is $0.5 \text{ m}$) (in $\text{ m}$)

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