A student claims that he applied a force F = -k x on a particle,...

For a particle to execute simple harmonic motion, the restoring force must be of the form $F = -c(x - x_0)$, where $c > 0$ is a constant and $x_0 is the equilibrium position.<br><br>Given that the applied force is$F = -k\sqrt{x}$, we can equate the two expressions:<br><br>$-k\sqrt{x} = -c(x - x_0)$<br><br>$k = \dfrac{c(x - x_0)}{\sqrt{x}} = c \left(\sqrt{x} - \dfrac{x_0}{\sqrt{x}}\right)

To determine the variation of $k$ with respect to $x$, we differentiate $k$ with respect to $x$:

\dfrac{dk}{dx} = \dfrac{c}{2\sqrt{x}} + \dfrac{cx_0}{2x\sqrt{x}}

Since the particle executes SHM exclusively in the region $x > 0$, its equilibrium position $x_0$ must also be positive ($x_0 > 0$).

Given $c > 0$, $x > 0$, and $x_0 > 0$, we find that both terms in the derivative are positive. Thus, \dfrac{dk}{dx}$> 0$.

Therefore, as $x$ increases, $k$ must increase.