Answer to Question #89129 in Mechanics | Relativity for Shivam Nishad

Question

A body of mass 0.2 kg is suspended from a spring of force constant 100 Nm^-1. The frictional force acting on it is 6v Newton, Write dow: the equation of motion and calculate the period of free oscillations. If
a harmonic force F = 20 cos 20t is applied,
calculate the amplitude of forced oscillations.

Answer 1

In the absence of oscillations, equilibrium between the force of gravity and the restoring force takes place,${{F}_{r}}={{F}_{grav}}$ or $kx=mg$ (it is assumed that the spring is directed along the x axis). This gives the extension of the spring

$${{x}_{0}}=\frac{mg}{k}=\frac{0.2\,kg\cdot 10\,m\cdot {{s}^{-2}}}{100\,N\cdot {{m}^{-1}}}=0.02\,m$$

1) In order to obtain the equation of motion of the pendulum we write Newton's second law

$$ma={{F}_{el}}+{{F}_{fr}}$$

where $a$ is the acceleration, $a=\frac{{{d}^{2}}x}{d{{t}^{2}}}$ ; ${{F}_{r}}$ is the restoring force, ${{F}_{r}}=-kx$ ; ${{F}_{fr}}$ is the frictional force, ${{F}_{fr}}=-b\cdot v=-b\frac{dx}{dt}$ $(b=6\,N\cdot s\cdot {{m}^{-1}})$ . Then the equation of motion of the pendulum is

$$m\frac{{{d}^{2}}x}{d{{t}^{2}}}=-kx-b\frac{dx}{dt}$$

or

$$\frac{{{d}^{2}}x}{d{{t}^{2}}}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0$$

Solve the equation and find the period of oscillations. To find x we substitute into equation the solution as exponential function $x={{e}^{\lambda t}}$ . We get

$${{\lambda }^{2}}{{e}^{\lambda t}}+\frac{b}{m}\lambda {{e}^{\lambda t}}+\frac{k}{m}{{e}^{\lambda t}}=0$$

or

$$\left( {{\lambda }^{2}}+\frac{b}{m}\lambda +\frac{k}{m} \right){{e}^{\lambda t}}=0$$

Then we get the auxiliary equation

$${{\lambda }^{2}}+\frac{b}{m}\lambda +\frac{k}{m}=0$$

The solution of this equation is

$$\lambda =-\frac{b}{2m}\pm \sqrt{\frac{{{b}^{2}}}{4{{m}^{2}}}-\frac{k}{m}}=-\frac{b}{2m}\pm i\sqrt{\frac{k}{m}-\frac{{{b}^{2}}}{4{{m}^{2}}}}$$

where $\sqrt{\frac{k}{m}-\frac{{{b}^{2}}}{4{{m}^{2}}}}=\omega$  is the angular frequency. The period of free oscillations is

$$T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{k}{m}-\frac{{{b}^{2}}}{4{{m}^{2}}}}}$$

Substituting known values $m=0.2\,kg,\,\,\,k=100\,N\cdot {{m}^{-1}},\,\,\,b=6\,N\cdot s\cdot {{m}^{-1}}$ , we get $T\approx 0.38\,s$

2) If a harmonic force F = 20 cos 20t= F₀ cos Ωt  is applied, then Newton's second law is

$$m\frac{{{d}^{2}}x}{d{{t}^{2}}}=-kx-b\frac{dx}{dt}+F$$

Then the equation of motion of the pendulum is

$$\frac{{{d}^{2}}x}{d{{t}^{2}}}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=\frac{{{F}_{0}}}{m}\cos \Omega t$$

or

$$\frac{{{d}^{2}}x}{d{{t}^{2}}}+\alpha \frac{dx}{dt}+\omega _{0}^{2}x={{A}_{0}}\cos \Omega t$$

where $\alpha =b/m,\,\,\,\omega _{0}^{2}=b/m,\,\,\,{{A}_{0}}={{F}_{0}}/m$ are denoted.

To calculate the amplitude of forced oscillations we find the particular solution of this equation. We seek a particular solution of the form

$${{x}_{p}}\left( t \right)=A\text{cos }\!\!\Omega\!\!\text{ }t+B\text{sin }\!\!\Omega\!\!\text{ }t$$

where A and B are unknowing constants which we now find. Substitute ${{x}_{p}}\left( t \right)$ into equation

$$-A{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}\text{cos }\!\!\Omega\!\!\text{ }t-B{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}\text{sin }\!\!\Omega\!\!\text{ }t+\alpha \left( -A\text{ }\!\!\Omega\!\!\text{ sin }\!\!\Omega\!\!\text{ }t+B\text{ }\!\!\Omega\!\!\text{ cos }\!\!\Omega\!\!\text{ }t \right)$$

$$+\omega _{0}^{2}\left( A\text{cos }\!\!\Omega\!\!\text{ }t+B\text{sin }\!\!\Omega\!\!\text{ }t \right)={{A}_{0}}\text{cos }\!\!\Omega\!\!\text{ }t$$

or

$$\left( -A{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}+B\alpha \text{ }\!\!\Omega\!\!\text{ }+A\omega _{0}^{2} \right)\text{cos }\!\!\Omega\!\!\text{ }t$$

$$+\left( -B{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}-A\alpha \text{ }\!\!\Omega\!\!\text{ }+B\omega _{0}^{2} \right)\text{sin }\!\!\Omega\!\!\text{ }t={{A}_{0}}\text{cos }\!\!\Omega\!\!\text{ }t$$

This is true if

$$-A{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}+B\alpha \text{ }\!\!\Omega\!\!\text{ }+A\omega _{0}^{2}={{A}_{0}}$$

and

$$-B{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}-A\alpha \text{ }\!\!\Omega\!\!\text{ }+B\omega _{0}^{2}=0$$

Solving this system of equation we get

$$A=\frac{{{A}_{0}}\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}$$

$$B=\frac{{{A}_{0}}\alpha \text{ }\!\!\Omega\!\!\text{ }}{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}$$

Then the particular solution is

$${{x}_{p}}\left( t \right)={{A}_{0}}\left( \frac{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)\text{cos }\!\!\Omega\!\!\text{ }t}{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}+\frac{\alpha \text{ }\!\!\Omega\!\!\text{ sin }\!\!\Omega\!\!\text{ }t}{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}} \right)$$

Denote

$$\text{cos}\varphi =\frac{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}{\sqrt{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}},\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ sin}\varphi =\frac{\alpha \text{ }\!\!\Omega\!\!\text{ }}{\sqrt{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}}$$

then

$${{x}_{p}}\left( t \right)=\frac{{{A}_{0}}}{\sqrt{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}}\left( \text{cos}\varphi \text{cos }\!\!\Omega\!\!\text{ }t+\text{sin}\varphi \text{sin }\!\!\Omega\!\!\text{ }t \right)$$

or

$${{x}_{p}}\left( t \right)=\frac{{{A}_{0}}}{\sqrt{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}}\text{cos}\left( \text{ }\!\!\Omega\!\!\text{ }t-\varphi \right)$$

The amplitude of forced oscillations is

$${{x}_{f}}=\frac{{{A}_{0}}}{\sqrt{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}}$$

Substituting Ω=20,

$${{A}_{0}}=\frac{{{F}_{0}}}{m}=\frac{20}{0.2}=100,\,\,\,\omega _{0}^{2}=\frac{k}{m}=\frac{100}{0.2}=500$$

$$\alpha =\frac{b}{m}=\frac{6}{0.2}=30$$

we get

$${{x}_{f}}=\frac{{{A}_{0}}}{\sqrt{{{\left( \omega _{0}^{2}-{{\text{ }\!\!\Omega\!\!\text{ }}^{2}} \right)}^{2}}+{{\alpha }^{2}}{{\text{ }\!\!\Omega\!\!\text{ }}^{2}}}}=\frac{100}{\sqrt{{{\left( 500-400 \right)}^{2}}+900\cdot 400}}\approx 0.164\,meters$$

Answer:

the equation of motion of the pendulum is

$$\frac{{{d}^{2}}x}{d{{t}^{2}}}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x=0$$

the period of free oscillations is $T\approx 0.38\,s$

the amplitude of forced oscillations is ${{x}_{f}}\approx 0.164\,meters$