A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?

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Solution

$r = 10 c m$

Because, $K E = P . E .$

So, $(\frac{1}{2}) m ω^{2} (r^{2} - y^{2}) = (\frac{1}{2}) m ω^{2} y^{2}$

$r^{2} - y^{2} = y^{2}$

$2 y^{2} = r^{2}$

$\Rightarrow y = \frac{r}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5 \sqrt{2}$

from the mean position

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A particle executes simple harmonic motion with an amplitude of 10 cm. At what distance from the mean position are the kinetic and potential energies equal ?

r=10 cm Because, KE=P.E. So, (12)mω2 (r2−y2)=(12)mω2y2 r2−y2=y2 2y2=r2 ⇒ y=r√2=10√2=5√2 from the mean position

$r = 10 c m$

Because, $K E = P . E .$

So, $(\frac{1}{2}) m ω^{2} (r^{2} - y^{2}) = (\frac{1}{2}) m ω^{2} y^{2}$

$r^{2} - y^{2} = y^{2}$

$2 y^{2} = r^{2}$

$\Rightarrow y = \frac{r}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5 \sqrt{2}$

from the mean position