Question

A mass oscillates along the x-axis according to the law, x = x0 cos $\left( \omega t - \frac{\pi}{4} \right)$. If the acceleration of the particle is written as a = A cos ($\omega t + \delta$), then

• A. A = x0$\omega$2, $\delta$ =$\frac{3\pi}{4}$
• B. A = x0, $\delta$ = $- \frac{\pi}{4}$
• C. A = x0$\omega$2, $\delta$ =$\frac{\pi}{4}$
• D. A = x0$\omega$2, $\delta$ =$- \frac{\pi}{4}$

Answer 1

Answer: (A)

A = x0$\omega$2, $\delta$ =$\frac{3\pi}{4}$

Answer 2

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

Velocity, $v = \frac{dx}{dt} = - x_{0}\omega\sin\left( \omega t - \frac{\pi}{4} \right)$

Acceleration, $a = \frac{dv}{dt} = - x_{0}\omega^{2}\cos\left( \omega t - \frac{\pi}{4} \right)$

$= x_{0}\omega^{2}\cos\lbrack\pi + (\omega t - \frac{\pi}{4})\rbrack$

$= x_{0}\omega^{2}\cos\lbrack\omega t + \frac{3\pi}{4}\rbrack$

Comparing it with acceleration

$a = A\cos(\omega t + \delta),$ we getr5

$A = x_{0}\omega^{2},\delta = \left( \frac{3\pi}{4} \right)$