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A mass of Hg is executing SHM which is given by $x=6.0 \cos \left(100 t+\frac{\pi}{4}\right) \mathrm{cm}$. What is the maximum kinetic energy?
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The correct answer is:
$18 J$
Here, $\mathrm{m}=1 \mathrm{~kg}$
The given equation of SHM is
$x=6.0 \cos \left(100 t+\frac{\pi}{4}\right)$
Comparing it with equation of SHM,
$\begin{aligned} x & =A \cos (\omega t+\phi), \text { we have } \\ A & =6.0 \mathrm{~cm}=\frac{6}{100} \mathrm{~m}\end{aligned}$
and $\quad \omega=100 \mathrm{rad} / \mathrm{s}$
Maximum kinetic energy $=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\max }\right)^2$
$=\frac{1}{2} \mathrm{~m}(\mathrm{~A} \omega)^2=\frac{1}{2} \times 1 \times\left[\frac{6}{100} \times 100\right]^2=18 \mathrm{~J}$
The given equation of SHM is
$x=6.0 \cos \left(100 t+\frac{\pi}{4}\right)$
Comparing it with equation of SHM,
$\begin{aligned} x & =A \cos (\omega t+\phi), \text { we have } \\ A & =6.0 \mathrm{~cm}=\frac{6}{100} \mathrm{~m}\end{aligned}$
and $\quad \omega=100 \mathrm{rad} / \mathrm{s}$
Maximum kinetic energy $=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\max }\right)^2$
$=\frac{1}{2} \mathrm{~m}(\mathrm{~A} \omega)^2=\frac{1}{2} \times 1 \times\left[\frac{6}{100} \times 100\right]^2=18 \mathrm{~J}$
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