A spring executes SHM with a mass of 10,kg at | vedclass

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(B) Given: Mass $m = 10\,kg$,Force constant $k = 10\,N/m$,Velocity $v = 40\,cm/s = 0.4\,m/s$,Amplitude $A = 0.5\,m$.First,calculate the angular freque

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$0.3$

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(B) Given: Mass $m = 10\,kg$,Force constant $k = 10\,N/m$,Velocity $v = 40\,cm/s = 0.4\,m/s$,Amplitude $A = 0.5\,m$.First,calculate the angular frequency $\omega$:$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{10}{10}} = 1\,rad/s$.The velocity $v$ at displacement $y$ is given by the formula:$v = \omega \sqrt{A^2 - y^2}$.Substituting the values:$0.4 = 1 	imes \sqrt{(0.5)^2 - y^2}$.Squaring both sides:$(0.4)^2 = (0.5)^2 - y^2$.$0.16 = 0.25 - y^2$.Solving for $y^2$:$y^2 = 0.25 - 0.16 = 0.09$.Taking the square root:$y = \sqrt{0.09} = 0.3\,m$.

$A$ spring executes $SHM$ with a mass of $10\,kg$ 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 will be .... $m$ (where amplitude is $0.5\,m$).

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