Question

For a particle executing $SHM$ along a straight line, match the statements in column$-I$ with statement in column $-II$. (Note that displacement given is column $-I$ is to be measured from mean position).

Answer 1

For SHM, we know $a=-kx$

we can write $a=v\frac{dv}{dx}$

$v\frac{dv}{dx}=-kx$

$vdv=-kxdx$

integrating both side:

$v^2=-k{x}^2$

$v\propto x$ i.e velocity displacement graph is straight line.

as we derived $v\propto x$ and

we know $a\propto x$

so $a\propto v$ acceleration-velocity graph is straight line.

it is known that, $a=-kx$; so acceleration-displacement graph is straight line.

it is known that in SHM, $x\propto sinwt$ and $a=-kx$;

so $a\propto sinwt$, acceleration-time graph is a sin function