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A heavy small-sized sphere is suspended by a string of length $l$. The sphere rotates uniformly in a horizontal circle with the string making an angle $\theta $ with the vertical. Then the time period of this conical pendulum is
$t = 2\pi \sqrt {\frac{l}{g}} $
$t = 2\pi \sqrt {\frac{{l\,\sin \,\theta }}{g}} $
$t = 2\pi \sqrt {\frac{{l\,\cos \,\theta }}{g}} $
$t = 2\pi \sqrt {\frac{l}{{g\,\cos \,\theta }}} $
Solution
Radius of circular path in the horizontal plane
$\mathrm{r}=l \sin \theta$
Forces acting on the bob are.
$(i)$ $\mathrm{T}=$ tension in the string
$(ii)$ $\mathrm{Mg}=$ weight of the bob
Resolving $T$ along the vertical and horizontal
direction, we get
${\mathrm{T} \cos \theta=\mathrm{Mg}}$ $…(1)$
${\mathrm{T} \sin \theta=\mathrm{Mr} \omega^{2}=\mathrm{M}(l \sin \theta) \omega^{2}}$
or ${\mathrm{T}=\mathrm{M} l \omega^{2}}$ $…(2)$
Dividing eq. $(2)$ by eq. $(1)$, we get
$\frac{1}{\cos \theta}=\frac{l \omega^{2}}{g}$
$\mathbf{o r}$ $\omega^{2}=\frac{g}{l \cos \theta}$
$\text { time period } \quad \mathrm{t}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{l \cos \theta}{\mathrm{g}}}$
