Question

In SHM, the distance of particle from middle point of its path at three consecutive seconds are found to be $x,y$ and $z$. Then time period of SHM is:

• A. $2\pi {\cos }^{-1}\left(\frac{x+z}{2y}\right)$
• B. $2\pi {\tan }^{-1}\left(\frac{x+y}{z}\right)$
• C. $\frac{2\pi }{{\sin }^{-1}\left(\frac{x+z}{2y}\right)}$
• D. $\frac{2\pi }{{\cos }^{-1}\left(\frac{x+z}{2y}\right)}$

Answer 1

The correct option is D $\frac{2\pi }{{\cos }^{-1}\left(\frac{x+z}{2y}\right)}$

$x=A\sin \omega t$
At $t=1,x=A\sin \omega$
At $t=2,y=A\sin 2\omega$
At $t=3,z=A\sin 3\omega$
$\therefore \frac{x+z}{2y}=\frac{A[\sin \omega +\sin 3\omega ]}{2A\sin 2\omega }$
$=\frac{2\sin 2\omega .\cos \omega }{2\sin 2\omega }=\cos \omega$
$\therefore \omega ={\cos }^{-1}\left(\frac{x+z}{2y}\right)$