Question

A particle moves with simple harmonic motion along x-axis. At times $t$ and $2t$, its positions are given by $x=a$ and $x=b$ respectively from equilibrium position. Find the time period of oscillation.

Answer 1

If initially particle is at

mean position. Then,

$x=Asin\left(\omega t\right)$ is the expression

of its displacement from mean. So,

$a=Asin\left(\omega t\right)$ ____ (i)

$b=Asin\left(2\omega t\right)$ ____ (ii)

Dividing (ii) by (i), we get

$\frac{sin\left(2cos\right)}{sin\left(\omega t\right)}=\frac{b}{a}$

$\frac{2sin\left(\omega t\right)cos\left(\omega t\right)}{sin\left(\omega t\right)}=\frac{b}{a}$

$\Rightarrow cos\left(\omega t\right)=\frac{b}{2a}$

$\Rightarrow \omega t=co{s}^{-1}\left(\frac{b}{2a}\right)$

As $\omega =2\pi f\Rightarrow \omega =\frac{2\pi }{1}$

So, $\frac{2\pi t}{T}=co{s}^{-1}\left(\frac{b}{2a}\right)$

$\Rightarrow T=\frac{2\pi t}{co{s}^{-1}\left(\frac{b}{2a}\right)}$