Question

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

Answer 1

Answer: (D)

Here, displacement $x=x_0\cos (\omega t-\pi /4)$

Velocity, $v=\frac{dx}{dt}=-x_0\omega \sin (\omega t-\pi /4)$

Acceleration, $a=\frac{dv}{dt}=-x_0{\omega }^2\cos (\omega t-\pi /4)$

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

or $a=x_0{\omega }^2\cos (\omega t-3\pi /4)....\left(1\right)$

Also given, $a=A\cos (\omega t+\delta )...\left(2\right)$

Comparing (1) and (2), we get $A=x_0{\omega }^2$ and $\delta =\frac{3\pi }{4}$