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JEE Mains · Physics · STD 11 - 13. oscillations

A pendulum bob has a speed of \(3\, {m} / {s}\) at its lowest position. The pendulum is \(50 \,{cm}\) long. The speed of bob, when the length makes an angle of \(60^{\circ}\) to the vertical will be \( .......\,{m} / {s}\) \(\left(g=10 \,{m} / {s}^{2}\right)\)

  1. A \(1\)
  2. B \(20\)
  3. C \(40\)
  4. D \(2\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(2\)

Step-by-step Solution

Detailed explanation

Applylng work energy theorem \(w _{ g }+ w _{ T }=\Delta K\) \(- mgl \left(1-\cos 60^{\circ}\right)=\frac{1}{2} mv ^{2}-\frac{1}{2} mu ^{2}\) \(v ^{2}= u ^{2}-2 gl \left(1-\cos 60^{\circ}\right)\) \(v ^{2}=9-2 \times 10 \times 0.5\left(\frac{1}{2}\right)\) \(v ^{2}=4\)…
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