WBJEE · Physics · Work Power Energy
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 \(\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{~g}}\)
- B \(\frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}\)
- C \(\frac{v_{0}^{2}}{4 g}\)
- D \(\frac{9 v_{0}^{2}}{4 g}\)
Answer & Solution
Correct Answer
(C) \(\frac{v_{0}^{2}}{4 g}\)
Step-by-step Solution
Detailed explanation
Hint: \(T=2\) sec. at \(\mathrm{t}=\mathrm{T}+\mathrm{T} / 8\) \(\mathrm{t}=\mathrm{T}+\mathrm{T} / 8 \quad \mathrm{v}=\mathrm{A} \omega \cos \omega \mathrm{t}=\mathrm{v}_{0} / \sqrt{2}\) By Mechanical energy conservation:…
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