Answer to Question #85684 in Mechanics | Relativity for Bhagwan Das

Question

A spring is stretched 5 10 m
−2
× by a force of .N 5 10−4
× 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) Us the following expression:

$$x(t)=A\cdot \text{e}^{-\delta t}\cdot \text{sin}(\omega_{spring}t)+F(t)/k,$$

where

$$A=\frac{mg}{k},$$

$$\omega_{spring}=\sqrt{k/m},$$

$$\delta=\frac{\beta}{2m}=0,$$

because we have zero friction, and

$$k=\frac{F_{stretching}}{x_{stretching}}.$$

Substitution of the values gives

$$x(t)=9.8\cdot \text{sin}(t)+20000\cdot \text{cos}(\omega t).$$

ii) Resonance happens when the angular frequency of the driving force is the same as the angular frequency of the spring:

$$\omega_{resonance}=1 \text{ r}^{-1}.$$