A spring executes S.H.M. with mass 10 kg att | vedclass

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(A) The velocity $v$ of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega = \sqrt{\frac{k}{m}}$.Given: $m = 10 \ kg$,$k =

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$0.3$

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(A) The velocity $v$ of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega = \sqrt{\frac{k}{m}}$.Given: $m = 10 \ kg$,$k = 10 \ N/m$,$A = 0.5 \ m$,and $v = 40 \ cm/s = 0.4 \ m/s$.First,calculate the angular frequency $\omega = \sqrt{\frac{10}{10}} = 1 \ rad/s$.Now,substitute the values into the velocity equation: $0.4 = 1 \cdot \sqrt{(0.5)^2 - x^2}$.Squaring both sides: $0.16 = 0.25 - x^2$.Rearranging for $x^2$: $x^2 = 0.25 - 0.16 = 0.09$.Taking the square root: $x = 0.3 \ m$.

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

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