For any $S.H.M.$, amplitude is $6\, cm$. If instantaneous potential energy is half the total energy then distance of particle from its mean position is .... $cm$
A$3$
B$4.2$
C$5.8$
D$6$
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B$4.2$
b (b) If at any instant displacement is $y$ then it is given that $U = \frac{1}{2} \times E$ ==> $\frac{1}{2}m{\omega ^2}{y^2} = \frac{1}{2} \times \left( {\frac{1}{2}m{\omega ^2}{a^2}} \right)$
==> $y = \frac{a}{{\sqrt 2 }} = \frac{6}{{\sqrt 2 }} = 4.2\,cm$
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The equation of an $S.H.M.$ with amplitude $A$ and angular frequency $\omega$ in which all the distances are measured from one extreme position and time is taken to be zero at the other extreme position is ...
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