Question

When the string of a
conical pendulum makes
an angle of 45° with the
vertical, its time period is
T1 . When the string
makes an angle of 60°
with the vertical, its time
period is T2 . Then T1²/T2² will be

Ans. √2. Please explain.

Answer 1

may be the question is  

\$\frac{T_{2} ^{2}}{T_{1}^{2}  }$ = $\sqrt{2}$

T1= 2$\pi \sqrt{\frac{l}{gcos\alpha }$

where alpha is 45 in first case and 60 in second you get your answer

Hope it helps!!

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