Question

Figure gives the x-t plot of particle executing one-dimensional simple harmonic motion.Give the signs of position, velocity and acceleration variables of the particle at t=0.3 s, 1.2 s , -1.2 s.

Answer 1

From graph

Time period

$T=2sec$

$w=\frac{2\pi }{T}=\frac{2\pi }{2}=\pi$

$x=-A\sin wt$

$-A\sin \pi t$

velocity of particle

$V_x=\frac{dx}{dt}=-A_0\pi cos\pi t$

acc of particle

$V_x=\frac{dyx}{dt}=-A_0\pi \sin \pi t$

At $t=0.3t$sec

$x=-A_0\sin \pi \times \frac{3}{10}$

$=-A_0\sin \frac{3\pi }{10}$(-ve)

$V=-A_0\pi cos\frac{3\pi }{10}(-ve)$

$a=A_0{\pi }^2\sin \frac{3\pi }{10}(+ve)$

$t=1.2\sec$

$x=-A_0\sin \pi \times \frac{12}{10}=-A_0\sin (\pi +\frac{\pi }{5})$

$=A_0\sin \frac{\pi }{5}(+ve)$

$v=-A_0\pi \cos (\pi +\frac{\pi }{5})=A_{0}cos\frac{\pi }{5}$ (+ve)

$a=-A_0{\pi }^2\sin \frac{\pi }{5}(-ve)$

At $t=-1.2sec$

$x=-A_0\sin (\pi \times \frac{162}{10})=-A_0\sin \left(\frac{-6\pi }{5}\right)=A_0\sin (\pi +\frac{\pi }{5})$ (-ve)

$v=-A_0\pi \cos (\pi \times \frac{-6}{5})=-A_0\pi \cos (\pi +\frac{\pi }{5})$

$=A_0\pi \cos \left(\pi /5\right)\to (+ve)$

$Q=A_0{\pi }^2\sin \left(\frac{-6\pi }{5}\right)=-A_0{\pi }^2\sin (\pi +\frac{\pi }{5})(+ve)$