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

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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