Answer to Question #85523 in Mechanics | Relativity for Kavita

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

First, calculate the spring constant:

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

Calculate the angular frequency of the spring with the mass:

$$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{0.01}{0.01}}=1 \text{ s}^{-1}.$$

i) The position of the mass at any time is given by

$$x(t)=-A\text{sin}(\omega t)=-\frac{mg}{k} \text{sin}(\omega t)=-9.8\text{sin}(t) \text{ m}.$$

ii) The resonance occurs when the angular frequency of the external force equals the natural angular frequency $\omega$, that is why

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