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a body oscillates with SHM according to the equation x=5cos[(2πt)+π/4]. At t=1.5, its acceleration is |
Answer» Answer:Given:Equation of SHM has been provided. x = 5 cos[(2πt) + π/4] To find:ACCELERATION at t = 1.5 seconds. Concept:Acceleration is obtained by 2nd order Differentiation of the displacement function wrt time. Calculation:x = 5 cos[(2πt) + π/4] ∴ v = dx/dt => v = 5 × (-2π) sin[(2πt) + π/4] => v = (-10π) sin[(2πt) + π/4] Now acc = dv/dt = d²x/dt² => acc. = (-10π)×(2π) cos[(2πt) + π/4] => acc. = (-20π²) cos[(2πt) + π/4] Now putting VALUE of t = 1.5 secs. we get : ∴ acc. = (-20π²) cos[{2π(3/2)} + π/4] => acc. = (-20π²) cos[3π + π/4] => acc. = (-20π²) cos[13π/4] Value of cos(13π/4) = (-1/√2) => acc. = (-20π²) × (-1/√2) => acc. = (20√2π²)/2 => acc. = 10√2 π² => acc. = 10 × 1.414 × π × π => acc. = 139.57728 m/s² So final answer : |
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