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13.Oscillations
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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}$ $\left(g=10 \,{m} / {s}^{2}\right)$
A
$1$
B
$20$
C
$40$
D
$2$
(JEE MAIN-2021)
Solution
Applylng work energy theorem
$w _{ g }+ w _{ T }=\Delta K$
$- mgl \left(1-\cos 60^{\circ}\right)=\frac{1}{2} mv ^{2}-\frac{1}{2} mu ^{2}$
$v ^{2}= u ^{2}-2 gl \left(1-\cos 60^{\circ}\right)$
$v ^{2}=9-2 \times 10 \times 0.5\left(\frac{1}{2}\right)$
$v ^{2}=4$
$v =2 m / s$
Standard 11
Physics
