Question

A point mass oscillates along the $x-$axis according to the law $x=x_0\cos (\omega t-\pi /4)$. If the acceleration of the particle is written as $a=A\cos (\omega t+\delta )$, then

• 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 equation is,
$x=x_0\cos (\omega t-\pi /4)$
Equation of acceleration,
$a=\frac{d^{2}x}{d{t}^2}=-{\omega }^2{x}_0\cos (\omega t-\pi /4)$
$\Rightarrow a={\omega }^2{x}_0\cos (\omega t-\pi /4+\pi )$
$[\because -\cos \theta =\cos (\pi +\theta \left)\right]$
$a={\omega }^2{x}_0\cos (\omega t+\frac{3\pi }{4})$
Compare it with equation $a=A\cos (\omega t+\delta )$, we get
$A={\omega }^2{x}_0$ and $\delta =\frac{3\pi }{4}$
Hence, option $\left(d\right)$ is the correct answer.