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

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Negative, Negative, Positive (at $t=0.3 s$ ) Positive, Positive, Negative (at $t=1.2 s$ ) Negative, Positive, Positive (at $t=-1.2 s )$ For simple harmonic motion (SHM) of a particle, acceleration ( $a$ ) is given by the relation:

$a=-\omega^{2} x \omega \rightarrow$ angular frequency $\ldots \ldots \ldots \ldots \ldots$ (i)

$t=0.3 s$

In this time interval, $x$ is negative. Thus, the slope of the $x-t$ plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive. $t=1.2 s$

In this time interval, $x$ is positive. Thus, the slope of the $x-t$ plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative. $t=-1.2 s$

In this time interval, $x$ is negative. Thus, the slope of the $x -t$ plot will also be negative. since both $x$ and $t$ are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.

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