Question

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

$x=x_{0}cos(\omega t-\pi /4)$
$v=\frac{dx}{dt}=-x_0\omega sin(\omega t-\pi /4)$
$a=\frac{dv}{dt}=-x_0{\left(\omega \right)}^{2}cos(\omega t-\pi /4)$
$=+x_0{\left(\omega \right)}^{2}cos[\pi +(\omega t-\pi /4)]$
$=x_0{\left(\omega \right)}^{2}cos(\omega t+3\pi /4)$
Comparing it with given equation,
$a=a_{0}cos(\omega t+\delta )$
$\Rightarrow a_0=x_0{\omega }^2$ and $\delta =\frac{3\pi }{4}$