1.

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.

\boxed{ \huge{ \red{acc. =  \dfrac{ {d}^{<klux>2</klux>}x }{ d {t}^{2}  }}}}

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 :

\boxed{\red{acc. \:= \: 139.57 \: ms^{-2}}}



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