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

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{π}{8} \sec$

$2 \tan^{- 1} \frac{4}{3} s$

$\frac{1}{4} \tan^{- 1} \frac{3}{4} s$

$\frac{1}{2} \tan^{- 1} \frac{3}{4} s$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed 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.

$∴ \sin (2 t - \tan^{- 1} \frac{3}{4}) = 0 \Rightarrow 2 t - \tan^{- 1} \frac{3}{4} = 0 \Rightarrow t = \frac{1}{2} \tan^{- 1} \frac{3}{4} \sec$

Watch 3-min video & get full concept clarity