Question

The bob of a swinging seconds pendulum (one whose time period is \(2 \mathrm{~s}\) ) has a small speed \(\mathrm{v}_{0}\) at its lowest point. Its height from this lowest point \(2.25 \mathrm{~s}\) after passing through it is given by
(A) \(\frac{v_{0}^{2}}{2 g}\)
(B) \(\frac{v_{0}^{2}}{g}\)
(C) \(\frac{v_{0}^{2}}{4 g}\)
(D) \(\frac{9 v_{0}^{2}}{4 g}\)

Answer 1

Ans: (C)
 \(\mathrm{T}=2\) sec.
at \(t=T+T / 8\)
\(t=T+T / 8\)
\(v=A \omega \cos \omega t=v_{0} / \sqrt{2}\)
By Mechanical energy conservation:
\(\left(\frac{v_{0}}{\sqrt{2}}\right)^{2}=v_{0}^{2}-2 g h\)
\(h=\frac{v_{0}^{2}}{4 g}\)

Answer 2

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Answer 3

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