Answer to Question #88572 in Classical Mechanics for Shivam Nishad

Question

A spring-mass system with m = 0.01 kg,
spring constant k = 36 Nm-1 and damping
constant y = 0.5 kg s-1 is subject to a
harmonic driving force. Calculate its
natural frequency and the resonance
frequency. How do they compare ? Discuss
the physical significance of your
result.

Answer 1

2nd Newtons law:

$$ma = -kx-yv \Rightarrow m\ddot{x}+y\dot{x}+kx=0$$

Characteristic equation:

$$m\lambda^2+y\lambda+k=0 \rightarrow \lambda = \frac{-y\pm i\sqrt{4 k m - y^2}}{2 m}$$

Therefore, natural frequency

$$\omega = \frac{\sqrt{4 k m - y^2}}{2 m}=\sqrt{ \frac{k}{m} - \frac{y^2}{4m^2}} = 54.5436\, \text{s}^{-1}$$

Assume external force with amplitude A and frequency $\Omega$. 2nd Newtons law takes the form

$$m\ddot{x}+y\dot{x}+kx=A\sin{\Omega t}$$

Solution:

$$x(t)=A\frac{ \sin (t \Omega +\phi )}{\sqrt{(k-m\Omega^2)^2 + y^2\Omega^2}}+\text{homogeneous eq. solution}$$

Resonance occurs at

$$\Omega = \sqrt{k/m}=60\, \text{s}^{-1}$$

Resonance frequency greater than natural freq. and doesn't depend on damping constant.