Physics - SHM Question with Solution | TestHub

The given equation of motion is $a=-bx$.

The standard equation for simple harmonic motion is $a=-{\omega }^{2}x$.

Comparing the two equations, we get ${\omega }^2=b$, which gives $\omega =\sqrt{b}$.

The periodic time $T$ is given by $T=\frac{2\pi }{\omega }$.

Substituting the value of $\omega$, we get $T=\frac{2\pi }{\sqrt{b}}$.

Answer: $\frac{2\pi }{\sqrt{\mathrm{b}}}$