Answer to Question #85565 in Mechanics | Relativity for Rinku Rai

Question

A spring is stretched 5 x10^-2 m by a force of 5x 10^−4 N. A mass of 0.01 kg is placed
on the lower end of the spring. After equilibrium has been reached, the upper end of the
spring is moved up and down so that the external force acting on the mass is given by
F(t) = 20 cos ωt. Calculate (i) the position of the mass at any time, measured form the
equilibrium position and (ii) the angular frequency for which resonance occurs.

Answer 1

i) The common oscillations (when the spring oscillates with the mass under the driving force influence) are described by the equation

$$x(t)=A\text{sin}(\omega_1 t)+F(t),$$

$$x(t)=-\frac{mg}{k} \text{sin}(\sqrt{ \frac{k}{m}} \cdot t) +F(t),$$

where

$$k=\frac{F_0}{x_0}=\frac{5\cdot 10^{-4}}{5\cdot 10^{-2}}=0.01\text{ N/m}.$$

Thus

$$x(t)=-9.8\text{sin}(t)+20\text{cos}(\omega t).$$

ii) The resonance occurs when the angular frequency of the driving force is equal to the angular frequency of the spring with the mass, i.e.

$$\omega_0=1 \text{ s}^{-1}$$