Question

A particle is moving with simple harmonic motion of time period $\left(\frac{2\pi }{n}\right)$ along a line . Amplitude a and centre of oscillation O. When t = $\frac{\pi }{24}$ it is at a distance $\left(\frac{\sqrt{3}}{2}\right)$ from O. The possible value of n is

Answer 1

The equation of SHM can be given by,

$x=Asin\omega t,$ where $\omega =\frac{2\pi }{T}$[T = Time period]

$=\frac{2\pi }{2\pi /n}=n$

so , x = A sin (nt)

when t = $\pi /24,x=\frac{\sqrt{3}}{2}$

$\frac{\sqrt{3}}{2}=Asin\left(\frac{n\pi }{24}\right)\Rightarrow \frac{n\pi }{24}=sin\left(\frac{\sqrt{3}}{2}\right)$

Now $si{n}_{-1}\left(x\right)$ can give values ranged in $[-\pi /2,\pi /2]$

also considering $x\le A,$ we get:-

$-\pi /2\le si{n}^{-1}\left(\frac{\sqrt{3}}{2A}\right)⩽+\pi /2$

$\Rightarrow -\pi /2⩽\frac{n\pi }{24}⩽\pi /2\Rightarrow -12⩽n⩽12.$ (Ans)