Question

A 10 kg mass stretches a spring 70 cm in equilibrium. Suppose a 2 kg mass is attached to the spring, initially displaced 25 cm below equilibrium, and given an upward velocity of 2 m/s. Find its displacement for t > 0. Find the frequency, period, amplitude, and phase angle of the motion

Answer 1

Answer:

Explanation:

1. Find spring constant k. From a free body diagram, you will get the forces on  the 10 kg mass, with a displacement d. It will be gravity pulling the mass down and the spring force pulling the mass up. The 10 kg mass is in equilibrium. The resulting equation will be:

$F = m_1a = kd - m_1g = 0 => m_1g = kd => k = \frac{m_1g}{d}$

2. Use the result from 1. to find the equations of motion. In general they are given by:

$x(t) = Acos(\omega t + \phi), v(t) = -\omega Asin(\omega t + \phi)$, where ω is:$\omega = \sqrt{\frac{k}{m_2}}=\sqrt{\frac{m_1g}{m_2d}}$

To find the amplitude A and the phase angle Ф, use the given initial conditions:

m₂ = 2 kg, x(0) = -0.25 m, v(0) = 2m/s

$-0.25 = Acos(\phi)\\2 = -\omega Asin(\phi)= -\sqrt{\frac{m_1g}{m_2d}}Asin(\phi)\\-0.24 = Asin(\phi)$

Solving for Ф:

$\frac{0.24}{0.25}=tan(\phi) => \phi = 0.76$

Solving for A:

$-0.25 = Acos(\phi) => A = -\frac{0.25}{cos(\phi)}=-0.35$

The equation for x(t) is now:

$x(t) = -0.35 cos(\sqrt{\frac{m_1g}{m_2d}t} +0.76)$

The frequency f is given by:$f = \frac{\omega}{2\pi}$

The period T is given by:$T = \frac{2\pi}{\omega}$