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

(D) $A = x_0\omega^{2},\ \delta = \dfrac{3\pi}{4}$

(A) $A = x_0,\ \delta = -\dfrac{\pi}{4}$

(B) $A = x_0\omega^{2},\ \delta = -\dfrac{\pi}{4}$

(C) $A = x_0\omega^{2},\ \delta = -\dfrac{\pi}{4}$

JEE Main · 2007 · 29 Apr 2007 · Offline · Elasticity & SHM

A point mass oscillates along the x-axis according to the law

x = x_{0} cos (ω t - \frac{π}{4})

. If the acceleration of the particle is written as

a = A cos (ω t + δ)

, then

Answer: (D)

Differentiating

x = x_{0} cos (ω t - \frac{π}{4})

twice gives the acceleration

a = - x_{0} ω^{2} cos (ω t - \frac{π}{4})

. Using

- cos θ = cos (θ + π)

, rewrite this as

a = x_{0} ω^{2} cos (ω t - \frac{π}{4} + π) = x_{0} ω^{2} cos (ω t + \frac{3 π}{4})

. Comparing with

a = A cos (ω t + δ)

identifies the amplitude as

A = x_{0} ω^{2}

and the phase as

δ = \frac{3 π}{4}

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option