Question

A point mass oscilllates along the x -axis according to the law x = x₀cos(ωt -π/4).if the acceleration of the particle is written as a = Acos(ωt+δ) then ,
(AIEEE 2007)

(a) A = x₀ , δ = -π/4
(b) A = x₀ω² , δ = π/4
(c) A = x₀ω² , δ = -π/4
(d ) A = x₀ω² , δ = 3π/4 ​

Answer 1

Answer:

(c) A = x₀ω² , δ = -π/4

Explanation:

x = x₀cos(ωt -π/4)

by differentiating with respect to x we will get velocity,

v = x₀x = x₀ωsin(ωt -π/4)

and again differentiation with respect to x we will get acceleration,

a = x₀ω²cos(ωt -π/4)

So, comparing with the given equation,

a = Acos(ωt+δ)

A = x₀ω²

δ = -π/4

Answer 2

Explanation:- Given equation of oscillation x = x₀cos(ωt -π/4) and general equation y = Acos(ωt±δ)Acceleration ,

a = d²x/dt² = d²(Acos(ωt-π/4)/dt²

• = -ω²x₀cos(ωt-π/4)
• = ω²x₀cos(ωt+3π/4)

compair with general equation .

• A = ω²x₀
• And , δ = 3π/4