A spring executes $SHM$ with mass of $10\,kg$ attached to it. The force constant of spring is $10\,N/m$.If at any instant its velocity is $40\,cm/sec$, the displacement will be .... $m$ (where amplitude is $0.5\,m$)
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(b) Angular velocity $\omega = \sqrt {\left( {\frac{k}{m}} \right)} $$ = \sqrt {\left( {\frac{{10}}{{10}}} \right)} = 1$
Now $u = \omega \sqrt {{a^2} - {y^2}} $
Now $u = \omega \sqrt {{a^2} - {y^2}} $
==>${y^2} = {a^2} - \frac{{{u^2}}}{{{\omega ^2}}}$$ = {(0.5)^2} - \frac{{{{(0.4)}^2}}}{{{1^2}}}$
==> ${y^2} = 0.9$
$y = 0.3\,m$
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