Question

A conical pendulum of length L makes an angle $\theta$ with the vertical. The time period will be

• A. $2\pi \sqrt{\frac{Lcos\theta }{g}}$
• B. $2\pi \sqrt{\frac{L}{gcos\theta }}$
• C. $2\pi \sqrt{\frac{Ltan\theta }{g}}$
• D. $2\pi \sqrt{\frac{L}{gtan\theta }}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{L}{gcos\theta }}$

Let T be the tension in string and m be the mass of pendulum bob.
Taking components of forces acting, we get



$Tcos\theta =mg$ ------- (i)
$Tsin\theta =mR{\omega }^2$ ------ (ii)

On dividing (ii)/(i)
$\frac{R{\omega }^2}{g}=tan\theta$
$\omega =\sqrt{\frac{gtan\theta }{R}},R=Lsin\theta$
$T=2\pi \sqrt{\frac{L}{gcos\theta }}$