Question

Consider a particle moving in simple harmonic motion according to the equation $x=2.0cos(50\Pi t+ta{n}^{-1}0.75)$

where x is in centimeter and t in second. The motion is started at t = 0. (a) When does the particle come to rest for the second time?

• A. $3.6\times {10}^{-2}$ sec
• B. $1.6\times {10}^{-2}$ sec
• C. $4.8\times {10}^{-2}$ sec
• D. $6.4\times {10}^{-2}$ sec

Answer 1

The correct option is A

$3.6\times {10}^{-2}$ sec

$x=2.0cos(50\Pi t+ta{n}^{-1}0.75)$

$v=\frac{dx}{dt}=-\pi sin(50\Pi t+ta{n}^{-1}0.75)$

Find t when v=0

$\Rightarrow =-100\pi sin(50\pi t+ta{n}^{-1}0.75)$

$sin\theta =0$

$\theta =n\pi$

$\theta =\left(\right(50\pi t+ta{n}^{-1}0.75)=n\pi$

$t=\frac{n\pi -ta{n}^{-1}0.75}{50\pi }$ $(ta{n}^{-1}0.75=0.64)$

First time particle will come to rest after start (i.e., t=0) is when $\pi =1$. second time it will happen when n

$=2$

$t=\frac{2\pi -0.64}{50\pi }(\because ta{n}^{-1}0.75=0.64)$

$n=2t=\frac{2\pi -0.64}{50\pi }$

$=3.6\times {10}^{-2}sec$