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Two simple harmonic motions $x_{1} = 2 c o s 100 π t a n d x_{2} = 2 s i n 100 π t$ are superimposed on a particle. If $x_{1} a n d x_{2}$ are in cm and t is in second, the maximum velocity of the particle is (in m/s)

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$4 \sqrt{2} π$

$\sqrt{2} π$

$2 \sqrt{2} π$

$2 π$

answer is C.

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Detailed Solution

$x = x_{1} + x_{2} = (2 c o s 100 π t + 2 s i n 100 π t) c m$

$\Rightarrow x = 2 \sqrt{2} c o s (100 π t - \frac{π}{4}) c m$

$∴ V_{\max} = Aω = 2 \sqrt{2} x 100 π cm / s = 2 \sqrt{2} π m / s$

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