a point mass oscillates along the x-axis according to the law x=xₒcos (wt -π/4).if the acceleration of the particle is written as a = A cos(wt + δ) then A = and δ = ?

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Given that x=xocosωt-π4so v=dxdt=-xoωsinωt-π4again differentiatinga=dvdt=-xoω2cosωt-π4a=-xoω2cosωt-π4absorbing the negative sign in cosine term ascosωt-π4+π=-cosωt-π4so -cosωt-π4=cosωt+3π4a=xoω2cosωt+3π4Thus on comparison A=xoω2    and δ=3π4

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Hope you get it

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wrong answer
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answer is A=x​ₒw² and δ = 3π/4
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Chandini if theta is -?/4 then it is 3?/4 only
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And magnitude of A is x 0 omega ^2
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