Question

A block of mass m is attached to a spring of spring constant k is free to oscillate with angular velocity $\omega$ in a horizontal plane without friction or clamping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. The amplitude of oscillations in terms of $\omega$, x0 and v0 is

• A. $\sqrt{\frac{\nu_{0}^{2}}{\omega^{2}} - x_{0}^{2}}$
• B. $\sqrt{\omega^{2}\nu_{0}^{2} + x_{0}^{2}}$
• C. $\sqrt{\frac{x_{0}^{2}}{\omega^{2}} + \nu_{0}^{2}}$
• D. $\sqrt{\frac{\nu_{0}^{2}}{\omega^{2}} + x_{0}^{2}}$

Answer 1

Answer: (D)

$\sqrt{\frac{\nu_{0}^{2}}{\omega^{2}} + x_{0}^{2}}$

Answer 2

Let the displacement of the block at instant of time t be

$x = A\cos(\omega t + \varphi)$

At $t = 0x = x_{0}$

$\therefore x_{0} = A\cos\varphi$

Velocity $v = \frac{dx}{dt} = - A\omega\sin(\omega t + \varphi)$

At $t = 0,v = - v_{0}$

$\therefore - v_{0} = - A\omega\sin\varphi$

Or $A\sin\varphi = \frac{v_{0}}{\omega}$ …… (i)

Squaring and adding (i) and (ii), we get

$A^{2}(\sin^{2}\varphi + \cos^{2}\varphi) = \frac{v_{0}^{2}}{\omega^{2}} + x_{0}^{2}$

$A = \sqrt{\frac{v_{0}^{2}}{\omega^{2}} + x_{0}^{2}}$