Question

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

• A. $A=x_0{\omega }^2,\delta =\frac{3\pi }{4}$
• B. $A=x_0,\delta =-\frac{\pi }{4}$
• C. $A=x_0{\omega }^2,\delta =\frac{\pi }{4}$
• D. $A=x_0{\omega }^2,\delta =-\frac{\pi }{4}$

Answer 1

The correct option is A $A=x_0{\omega }^2,\delta =\frac{3\pi }{4}$

$x=x_{0}cos(\omega t-\frac{\pi }{4})$

$v=\frac{dx}{dt}=-x_0\omega \sin (\cot -\frac{\pi }{4})$

$a=\frac{dv}{dt}=-x_0{\omega }^2\cos (\cot -\frac{\pi }{4})$

$\cos (\pi +\theta )=\theta$

$a=x_0{\omega }^2\cos (\omega t+\pi -\frac{\pi }{4})$

$a=x_0{\omega }^2\cos (\omega t+3\frac{\pi }{4})$

On comparing this equation with $a=A\cos (\omega t+\delta )$

$⟹A=x_0{\omega }^2$ and $\delta =\frac{3\pi }{4}$