Figure gives the x-t plot of a particle executing one dimensional simple harmonic motion. Mark the correct combination. The signs of position, velocity and acceleration variables at

Question

Figure gives the x-t plot of a particle executing one dimensional simple harmonic motion. Mark the correct combination. The signs of position, velocity and acceleration variables at (A) t=0.3 s , x< 0, v < 0, a < 0 (B) t=1.2 s , x > 0, v > 0, a >0 (C) t=-1.2 s , x > 0, v > 0, a < 0 (D) t=-0.3 s , x > 0, v < 0, a < 0

Tardigrade Admin · Accepted Answer

Between -0.5 < t < 0.5, the graph has a negative slope. ⇒ (d x/d t)<0 or v < 0 At t=-0.3 s , the slope becomes more and more negative, so (d v/d t) < 0 or a < 0 . x > 0 is visible.

Q. Figure gives the $x - t$ plot of a particle executing one dimensional simple harmonic motion. Mark the correct combination. The signs of position, velocity and acceleration variables at

$t = 0.3 s, x < 0, v < 0, a < 0$

$t = 1.2 s, x > 0, v > 0, a > 0$

$t = - 1.2 s, x > 0, v > 0, a < 0$

$t = - 0.3 s, x > 0, v < 0, a < 0$

Between $- 0.5 < t < 0.5$ , the graph has a negative slope.
$\Rightarrow \frac{d x}{d t} < 0$ or $v < 0$
At $t = - 0.3 s,$ the slope becomes more and more negative, so
$\frac{d v}{d t} < 0$
or $a < 0. x > 0$ is visible.

Q. Figure gives the x−t plot of a particle executing one dimensional simple harmonic motion. Mark the correct combination. The signs of position, velocity and acceleration variables at

t=0.3s,x<0,v<0,a<0

t=1.2s,x>0,v>0,a>0

t=−1.2s,x>0,v>0,a<0

t=−0.3s,x>0,v<0,a<0