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Question

A body executing $$S.H.M.$$ along a straight line has a velocity of $$3\ ms^{-1}$$ when it is at a distance of $$4\ m$$ from its mean position and $$4\ ms^{-1}$$ when it is at  a distance of $$3\ m$$ from its mean position. Its angular frequency and amplitude are:


A
2 rad s1 & 5 m
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B
1 rad s1 & 10 m
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C
2 rad s1 & 10 m
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D
1 rad s1 & 5 m
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Solution

The correct option is D $$1\ rad\ s^{-1}$$ & $$5\ m$$
Velocity $$v=\omega \sqrt{A^2-X^2}$$
$$3=\omega \sqrt{A^2-4^2}$$...(1)
$$4=\omega\sqrt{A^2-3^2}$$...(2)
$$\dfrac{3}{4} =\dfrac{\sqrt{A^2-4^2}}{\sqrt{A^2-3^2}}$$
$$\dfrac{9}{16} =\dfrac{A^2-16}{A^2-9}$$
$$16A^2-9A^2=16^2-9^2$$
$$7A^2=175$$
$$A^2=25$$
$$A=5$$
From (1), $$3=\omega \sqrt{5^2-4^2}$$
$$\omega=1 rad/sec$$


Physics

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