Question

Acceleration-time graph of a particle in SHM is as shown in figure. Match the following two columns.

Answer 1

(A)At $t_1$ acceleration is positive therefore, displacement will be negative.
Since $a=-{\omega }^{2}x$
A->3
(B) Since acceleration is zero. therefore, displacement is zero.
B->1
(C)Since, $v=\int adt$
Therefore, the area of $a-t$ curve from $t=0$ to $t=t_1$ is negative therefore,
$v<0$
Alternatively, Just before, $t_1$ the point where $a-t$ curve crosses t-axis is making acceleration negative to positive and also the$a_{max}$ point is also after $t_1$ therefore, particle has just crosses the mean position and it is traveling in $-ve$ direction therefore, velocity is negative.
C->3
(D)$t_2$ is mean position therefore, x=0 and acceleration is zero but acceleration is becoming -ve to +ve therefore, displacement is becoming +ve to -ve therefore, particle is at mean position and travelling in -ve direction therefore, velocity is -ve and maximum.
D->3,4