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प्रश्न
The displacement of a particle is 'y' = 2 sin `[(pit)/2 + phi]`, where 'y' is cm and 't' in second. What is the maximum acceleration of the particle executing simple harmonic motion?
(Φ = phase difference)
पर्याय
`pi^2/4` cm/s2
`pi/2` cm/s2
`pi^2/2` cm/s2
`pi/4` cm/s2
MCQ
उत्तर
`underline(pi^2/2 cm"/"s^2)`
Explanation:
`y = 2sin((pit)/2 + phi)`
`dy/dt = (2pi)/2 cos((pit)/2 + phi)`
`(d^2y)/(dt^2) = (1)pi xx pi/2sin((pit)/2 + phi)`
∴ `(d^2y)/(dt^2)|_{max} = (-)pi^2/2` cm/s2
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Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.
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