AP EAMCET · PHYSICS · Oscillations
A bob of a pendulum of length \(0.5 \mathrm{~m}\) has a speed of \(6 \mathrm{~ms}^{-1}\) at its lowest point. Find the speed of the bob when the string of the pendulum makes \(60^{\circ}\) with the vertical, (take \(\left.g=10 \mathrm{~ms}^{-2}\right)\)
- A \(26 \mathrm{~ms}^{-1}\)
- B \(\sqrt{31} \mathrm{~ms}^{-1}\)
- C \(13 \mathrm{~ms}^{-1}\)
- D \(1.3 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{31} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, length of pendulum, \(l=0.5 \mathrm{~m}\) Speed at lowest point, \(v_1=6 \mathrm{~m} / \mathrm{s}\) Angle made by string with vertical, \(\theta=60^{\circ}\) In \(\triangle O B C, \frac{O C}{O B}=\cos \theta=\cos 60^{\circ}\) \(\frac{l-h}{l}=\cos 60^{\circ}\)…
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