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Given that the equation of motion of a m...

Given that the equation of motion of a mass is `x = 0.20 sin (3,0 t) m` . Find the velocity and acceleration of the mass when the object is `5 cm` from its equilibrium position. Repeat for `x = 0`.

Text Solution

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The correct Answer is:
B, D
`omega = 3rad//s`
`A = 0.2m`
At `x = 5cm`
`v = +-omegasqrt(A^(2) - x^(2))`
`= +- 3.0sqrt((0.2)^(2) - (0.05)^(2))`
`= +- 0.58m//s`
`a = - omega^(2)x = - (3)^(2)(0.05)`
` = - 0.45m//s`
At `x = 0`
`v = +-omega A`
`= +- (3.0)(0.2)`
`= +- 0.6 m//s`
`a = - omega^(2)x = - (3)^(2)(0)`
`= 0`
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