Question

Particle moves such that its acceleration a is given by a = -bx, where x is the displacement from equilibrium position and b is a constant. the period of oscillation is

Answer 1

it is the shortest method

Answer 2

Answer:

The time period of oscillation is $\dfrac{2\pi}{\sqrt{b}}$

Explanation:

Given that,

The acceleration a = -bx...(I)

Where, x = displacement

We know that,

The general formula of the acceleration

$a=-\omega^2\times x$....(II)

Where, a = acceleration

x = displacement

$\omega$=angular velocity

On comparing the both equation of acceleration

$-bx=-\omega^2\times x$

$b =\omega^2$

$\omega =\sqrt{b}$

The time period of oscillation is

$T=\dfrac{2\pi}{\omega}$

Put the value of $\omega$

$T=\dfrac{2\pi}{\sqrt{b}}$

Hence, The time period of oscillation is $\dfrac{2\pi}{\sqrt{b}}$