Question

The equation of motion of a particle in S.H.M. is given by $a=-B\times x$ ,where a is the acceleration, B is a constant and x is displacement. The period of oscillation of the particle is:

• A. 2$\pi \sqrt{B}$
• B. 2$\pi /B$
• C. 2$\pi /\sqrt{B}$
• D. 2 $\pi B$

Answer 1

$a=-Bx$
so, $B={\omega }^2$ $(\therefore a=-{\omega }^{2}x)$
$T=\frac{2\pi }{\omega }$
$=\frac{2\pi }{\sqrt{B}}$
$so,timeperiodofoscillations=\frac{2\pi }{\sqrt{B}}$