A point mass oscillates along the x-axis according to the law x = x0 cos (ω t − π/4). If the acceleration of the particle is written as a = A cos(ω t + δ) then

Question

A point mass oscillates along the x-axis according to the law x = x0 cos (ω t − π/4). If the acceleration of the particle is written as a = A cos(ω t + δ) then (A) A = x0 , δ = −π/4 (B) A = x0 ω2, δ = −π/4 (C) A = x0 ω2 , δ = −π/4 (D) A = x0 ω2, δ = 3π/4

Tardigrade Admin · Accepted Answer

v = âx0Ï sin (Ït â π/4) a = âx0Ï2cos(ω t+π-(π/4)) a = A cos(Ït + Î´) A = x0Ï2; δ=(3π/4)

Q. A point mass oscillates along the $x - a x i s$ according to the law $x = x_{0} cos (ω t - π /4)$ . If the acceleration of the particle is written as
$a = A cos (ω t + δ)$ then

$A = x_{0}, δ = - π /4$

$A = x_{0} ω^{2}, δ = - π /4$

$A = x_{0} ω^{2}, δ = - π /4$

$A = x_{0} ω^{2}, δ = 3 π /4$

$v = - x_{0} ω s in (ω t - π /4)$
$a = - x_{0} ω^{2} cos (ω t + π - \frac{π}{4})$
$a = A cos (ω t + δ)$
$A = x_{0} ω^{2}; δ = \frac{3 π}{4}$

Q. A point mass oscillates along the x−axis according to the law x=x0​cos(ωt−π/4). If the acceleration of the particle is written as a=Acos(ωt+δ) then

A=x0​,δ=−π/4

A=x0​ω2,δ=−π/4

A=x0​ω2,δ=−π/4

A=x0​ω2,δ=3π/4

v=−x0​ωsin(ωt−π/4) a=−x0​ω2cos(ωt+π−4π​) a=Acos(ωt+δ) A=x0​ω2;δ=43π​

Q. A point mass oscillates along the $x - a x i s$ according to the law $x = x_{0} cos (ω t - π /4)$ . If the acceleration of the particle is written as
$a = A cos (ω t + δ)$ then

$A = x_{0}, δ = - π /4$

$A = x_{0} ω^{2}, δ = - π /4$

$A = x_{0} ω^{2}, δ = - π /4$

$A = x_{0} ω^{2}, δ = 3 π /4$

$v = - x_{0} ω s in (ω t - π /4)$
$a = - x_{0} ω^{2} cos (ω t + π - \frac{π}{4})$
$a = A cos (ω t + δ)$
$A = x_{0} ω^{2}; δ = \frac{3 π}{4}$