JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2021
- Medium
- Numerical
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 \( .......\,{m} / {s}\) \(\left(g=10 \,{m} / {s}^{2}\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
Applylng work energy theorem
\(w _{ g }+ w _{ T }=\Delta K\)
\(- mgl \left(1-\cos 60^{\circ}\right)=\frac{1}{2} mv ^{2}-\frac{1}{2} mu ^{2}\)
\(v ^{2}= u ^{2}-2 gl \left(1-\cos 60^{\circ}\right)\)
\(v ^{2}=9-2 \times 10 \times 0.5\left(\frac{1}{2}\right)\)
\(v ^{2}=4\)
\(v =2 m / s\)
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