Simple hormonic motion

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A particle is executing SHM along a straight line and its position at time t is given by $x = (4 c o s 2 t + 3 s i n 2 t)$ cm. Then the particle arrives at its extreme position after a time of

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Solution

$x = 4 \cos 2 t + 3 \sin 2 t = \sqrt{4^{2} + 3^{2}} . \cos (2 t - \tan^{- 1} (\frac{3}{4}))$
$∴ V = \frac{d x}{d t} = - 5 \times 2 \sin (2 t - \tan^{- 1} \frac{3}{4}) \frac{c m}{s}$
When the particle arrives at extreme position, its velocity becomes zero.
$∴ \sin (2 t - \tan^{- 1} \frac{3}{4}) = 0 \Rightarrow 2 t - \tan^{- 1} \frac{3}{4} = 0 \Rightarrow t = \frac{1}{2} \tan^{- 1} \frac{3}{4} \sec$

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Simple hormonic motion

Remember concepts with our Masterclasses.

A particle is executing SHM along a straight line and its position at time t is given by x=(4cos⁡2t+3sin⁡2t) cm. Then the particle arrives at its extreme position after a time of

x=4cos⁡2t+3sin⁡2t=42+32.cos(2t−tan⁡−1(34))∴V=dxdt=−5×2sin⁡(2t−tan⁡−134)cmsWhen the particle arrives at extreme position, its velocity becomes zero. ∴sin(2t−tan−134)=0⇒2t−tan−134=0⇒t=12tan−134 sec

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A particle is executing SHM along a straight line and its position at time t is given by $x = (4 c o s 2 t + 3 s i n 2 t)$ cm. Then the particle arrives at its extreme position after a time of