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A point mass oscillates along the x-axis according to the law $X = X_{0} \cos (ωt - π / 4)$ .If the acceleration of the particle is written as $α$ = $a = Acos (ωt + δ), then$

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a

$A = x_{0}, δ = - π / 4$

b

$A = x_{0} ω^{2}, δ = 3 π / 4$

c

$A = X_{0} ω^{2}, δ = - π / 4$

d

$A = X_{0} ω^{2}, δ = π / 4$

answer is B.

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Detailed Solution

$x = x_{0} \cos (ωt - \frac{π}{4}) a = \frac{d^{2} x}{{dt}^{2}} = - x_{0} ω^{2} \cos (ωt - \frac{π}{4}) = x_{0} ω^{2} \cos (ωt + \frac{3 π}{4}) Given a = Acos (ωt + δ) Comparing eqs . (1) and (2), we get A = x_{0} ω^{2} and δ = \frac{3 π}{4}$

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