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

Answer: (C)

$A=x_0{\omega }^2,\delta =-\pi /4$

$x=x_0\cos (\omega t-\frac{\pi }{4})$

$v={-x}_0\sin (\omega t-\frac{\pi }{4})X_w$

$a={-x}_0{\omega }^2\cos (\omega t-\frac{\pi }{4})$

Comparing this by

$a=A\cos (\omega t+\rho )$

we get, $A=x_0{\omega }^2$

and, $\delta =\frac{-\pi }{4}$

Thus option $C$ is correct.