Question

Find the period of small oscillation in a vertical plane performed by a ball of mass m=40 g fixed at the middle of a horizontally stretched string i=1.0 min length. The tension of the string is assumed to be constant and equal to F=10 N.

Answer 1

Let us locate and depict the forces acting on the ball at the position when it is at a distance $x$ down from the undeformed position of the string
As this position,the unbalanced downward force on the ball
$=mg-2F\sin \theta$
By Newton's law $m\ddot{x}=mg-2F\sin \theta$
$mg-2F\theta$ (when $\theta$ small)
$=mg-2F\frac{x}{l/2}=mg-\frac{4F}{l}x$
Thus $\ddot{x}=g-\frac{4F}{ml}x=-\frac{4F}{ml}(x-\frac{mgl}{4F})$
putting $x^{{}^{\prime }}=x-\frac{mgl}{T}$, we get
$\ddot{x}=\frac{4T}{ml}{x}^{{}^{\prime }}$
Thus $T=\pi \sqrt{\frac{ml}{F}}=0.2s$