Question

A point mass oscillates along the $x-axis$ according to the law $x = x_0 \,cos (\omega t − \pi/4)$. If the acceleration of the particle is written as
$a = A cos(\omega t + \delta)$ then

• A. $A = x_0 , \delta = −\pi/4$
• B. $A = x_0 \omega^2, \delta = −\pi/4$
• C. $A = x_0 \omega^2 , \delta = −\pi/4$
• D. $A = x_0 \omega^2, \delta = 3\pi/4$

Answer 1

Answer: (D)

$v = −x_{0}ω\, sin \left(ωt − \pi/4\right)$
$a = −x_{0}ω^{2}cos\left(\omega t+\pi-\frac{\pi}{4}\right)$
$a = A cos\left(ωt + δ\right)$
$A = x_{0}ω^{2}; \delta=\frac{3\pi}{4}$