What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of $10$ cm.(Take $g = 9.8 m / s^{2}$ )

0.6

m/s

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1.4

m/s

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1.8

m/s

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2.2

m/s

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Solution

The correct option is B $1.4$ m/s
KE at mean position= PE at maximum height
$\frac{1}{2} m v^{2} = m g h$
$v = \sqrt{2 g h}$

$= \sqrt{2 \times 9.8 \times 0.1}$

$= 1.4 \frac{m}{s}$

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What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of 10cm.(Take g=9.8m/s2)

The correct option is B 1.4 m/sKE at mean position= PE at maximum height12mv2=mghv=√2gh=√2×9.8×0.1 =1.4ms

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of $10$ cm.(Take $g = 9.8 m / s^{2}$ )

The correct option is B $1.4$ m/s
KE at mean position= PE at maximum height
$\frac{1}{2} m v^{2} = m g h$
$v = \sqrt{2 g h}$

$= \sqrt{2 \times 9.8 \times 0.1}$

$= 1.4 \frac{m}{s}$