The period of small oscillations of a simple pendulum of length - if its point of suspension O moves constant acceleration - - - 1 - - - 2 - with respect ti earth is

The correct option is A T = 2 π√(ℓ/{( g α _ 2 ) ^ 2 + α _ 1 ^ 2 } ^ 1 / 2 ) ...

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Standard XII

Physics

Physical Pendulum

The period of...

Question

The period of small oscillations of a simple pendulum of length $ℓ$ if its point of suspension O moves constant acceleration $α = α_{1}^i - α_{2}^j$ with respect ti earth is

T = 2 π \sqrt{\frac{ℓ}{{{(g - α_{2})}_{2}^{2} + α_{1}^{2}}^{1 / 2}}}

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T = 2 π \sqrt{\frac{ℓ}{{{(g - α_{1})}_{1}^{2} + α_{2}^{2}}^{1 / 2}}}

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T = 2 π \sqrt{\frac{ℓ}{g}}

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T = 2 π \sqrt{\frac{ℓ}{{(g^{2} + α_{1}^{2})}^{1 / 2}}}

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Solution

The correct option is A $T = 2 π \sqrt{\frac{ℓ}{{{(g - α_{2})}_{2}^{2} + α_{1}^{2}}^{1 / 2}}}$

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The period of small oscillations of a simple pendulum of length ℓ if its point of suspension O moves constant acceleration α=α1^i−α2^j with respect ti earth is

The correct option is A T=2π√ℓ{(g−α2)2+α21}1/2

The period of small oscillations of a simple pendulum of length $ℓ$ if its point of suspension O moves constant acceleration $α = α_{1}^i - α_{2}^j$ with respect ti earth is

The correct option is A $T = 2 π \sqrt{\frac{ℓ}{{{(g - α_{2})}_{2}^{2} + α_{1}^{2}}^{1 / 2}}}$