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Standard XII

Physics

Dimensional Analysis

A pendulum bo...

Question

A pendulum bob has a speed $3 m / s$ while passing through its lowest position, length of the pendulum is $0.5 m$ then its speed when it makes an angle of $60^{\circ}$ with the vertical is

2 m / s

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1 m / s

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4 m / s

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3 m / s

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Solution

The correct option is A $2 m / s$
Since energy is conserved:
$\frac{m u^{2}}{2} = \frac{m v^{2}}{2} + m g h \Rightarrow u^{2} = v^{2} + 2 g h \Rightarrow {(3)}^{2} = v^{2} + 2 (9.8) (0.5 - 0.5 cos 60) \Rightarrow v = 2 m / s$

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A pendulum bob has a speed 3m/s while passing through its lowest position, length of the pendulum is 0.5m then its speed when it makes an angle of 60∘ with the vertical is

The correct option is A 2m/sSince energy is conserved:mu22=mv22+mgh⇒u2=v2+2gh⇒(3)2=v2+2(9.8)(0.5−0.5cos60)⇒v=2m/s

A pendulum bob has a speed $3 m / s$ while passing through its lowest position, length of the pendulum is $0.5 m$ then its speed when it makes an angle of $60^{\circ}$ with the vertical is

The correct option is A $2 m / s$
Since energy is conserved:
$\frac{m u^{2}}{2} = \frac{m v^{2}}{2} + m g h \Rightarrow u^{2} = v^{2} + 2 g h \Rightarrow {(3)}^{2} = v^{2} + 2 (9.8) (0.5 - 0.5 cos 60) \Rightarrow v = 2 m / s$