Question

A point mass oscillates along the $X$ -axis according to the relation $x=x_{0}cos\left(\right.\omega t-\pi /4\left.\right).$ If acceleration of the particle is written as $a=a_{0}cos\left(\right.\omega t+\delta\left.\right),$ then

• A. $a_{0}=x_{0}\omega ^{2},\delta=-\pi /4$
• B. $a_{0}=x_{0}\omega ^{2},\delta=\pi /4$
• C. $a_{0}=x_{0}\omega ^{2},\delta=-3\pi /4$
• D. $a_{0}=x_{0}\omega ^{2},\delta=3\pi /4$

Answer 1

Answer: (D)

$x=x_{0}cos\left(\right.\omega t-\pi /4\left.\right)$
$v=\frac{d x}{d t}=-x_{0}\omega sin\left(\right.\omega t-\pi /4\left.\right)$
$a=\frac{d v}{d t}=-x_{0}\left(\omega \right)^{2}cos\left(\right.\omega t-\pi /4\left.\right)$
$=+x_{0}\left(\omega \right)^{2}cos\left[\right.\pi +\left(\right.\omega t-\pi /4\left.\right)\left]\right.$
$=x_{0}\left(\omega \right)^{2}cos\left(\right.\omega t+3\pi /4\left.\right)$
Comparing it with given equation,
$a=a_{0}cos\left(\right.\omega t+\delta\left.\right)$
$\Rightarrow a_{0}=x_{0}\omega ^{2}$ and $\delta=\frac{3 \pi }{4}$