A point mass oscillates along the $$x-axis$$ according to the law $$x=x0cos$$ ω$$t-$$π$$4$$. If the acceleration of the particle is written as $$a=A$$ $$cos$$ ω$$t+$$δ,then :

Correct option is 1A=xoω2,δ=3π4

Questions

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

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A=xoω2,δ=3π4

A=xo,δ=-π4

A=xoω2,δ=π4

A=xoω2,δ=-π4

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detailed solution

Correct option is A

Acceleration, a=d2xdt2=-x0ω2cosωt-π4 =x0ω2cos ωt-π4+π =x0ω2cosωt+3π4 ⇒ A=x0ω2and δ=3π4

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A point mass oscillates along the x-axis according to the law x=x0cos ωt-π4. If the acceleration of the particle is written as a=A cos ωt+δ,then :

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