Question

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

• 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

The correct option is D $A=X_0{\omega }^2,\delta =3\pi /4$

Given,

$X=X_{0}cos(\omega t-\pi /4)$

velocity, $v=\frac{dX}{dt}=-\omega X_{0}sin(\omega t-\pi /4)$

acceleration, $a=\frac{dv}{dt}=-X_0{\omega }^{2}cos(\omega t-\pi /4)$. . . . . .(1)

Now, we know that, $-cos\theta =cos(\pi +\theta )$ as displcament and acceleration should have $\pi$ phase difference

Therefore, $a=X_0{\omega }^{2}cos(\pi +(\omega t-\pi /4\left)\right)$

$\Rightarrow a=X_0{\omega }^{2}cos\left(\right(\omega t+3\pi /4\left)\right)$. . . . . . . . .(2)

acceleration, $a=Acos(\omega t+\delta )$. . . . . . .. .(3)

Compare equation (2) and (3), we get

$A=X_0{\omega }^2$ and $\delta =3\pi /4$

Option D is correct.