Individual tasks for PROBLEM 3.3

DAMPED HARMONIC OSCILLATIONS.

In accordance with your variant to solve one of the following problems listed below (The number of problem statement and all necessary input data are reduced in the table 3.3).

1 Load of mass of m, suspended on the spring with stiffness of k, oscillate in viscous medium with drag coefficient of r. The equation of oscillations of load has view x(t) = А₀·e^{- βt}·cosωt. Logarithmic decay decrement of oscillations is δ.

a) By the values of quantities, given in the table 3.3, find necessary parameters and write down the equation of vibrations with numerical coefficients.

b) Find the system quality factor.

c) Find the quantity, which is indicated in the last column of table.

2 The oscillation circuit consists of capacitor with capacity of С, coil of inductance of L and resistor of resistance of R. Current in circuit changes by law i(t) = I₀·e^{– βt}·sinωt. Logarithmic decay decrement of oscillations is δ.

a) By the values of quantities, given in the table 3.3, find necessary parameters and write down the equation of current’s oscillations with numerical coefficients.

b) Find the system quality factor.

c) Find the quantity, which is indicated in the last column of table.

TABLE OF TASK VARIANTS

Table 3.3

Variant

Statement

k , N/m

r , kg/s

m , g

А , cm

С , mF

L , mH

R , Ω

I0, mА

β , s–1

ω, rad/s

δ

Find

–

–

–

–

–

–

–

δ

–

–

–

–

–

–

–

0,9

L

–

–

–

–

–

–

–

1,2

С

–

–

–

–

–

–

–

104

L

–

0,5

–

–

–

–

–

–

k

–

–

–

–

–

–

–

1,7

С

–

–

–

–

–

–

–

1,1

r

–

–

–

–

–

–

–

1,8

L

–

–

–

–

–

–

–

1,5

С

–

–

–

–

–

–

–

1,6

k

–

–

–

–

0,5

–

–

–

δ

–

–

–

–

–

–

–

R

1,6

–

–

–

–

–

–

–

δ

–

–

–

–

–

–

–

1,9

С

–

–

–

–

–

–

–

1,8

r

–

–

–

–

–

–

5·103

–

δ

–

1,2

–

–

–

–

–

–

1,5

k

–

–

–

–

–

–

–

1,6

R

–

–

–

–

–

–

–

1,2

k

–

–

–

–

–

–

–

2·104

1,5

L

–

–

–

–

–

–

–

1,4

С

–

–

–

–

–

–

–

1,4

r

–

–

–

–

–

–

–

1,3

R

0,8

–

–

–

–

–

–

–

δ

–

–

–

–

0,2

–

–

–

1,1

L

–

–

–

–

–

–

–

1,3

k

–

–

–

–

–

–

–

1,1

С

–

1,1

–

–

–

–

–

–

1,4

k

–

–

–

–

–

–

–

δ

–

–

–

–

–

–

–

1,2

r

Problem 3.4.

DRIVEN HARMONIC OSCILLATIONS

MAIN CONCEPTS

Series oscillatory circuit.

Ohm’s law for the alternate current is represented the formula (11) together with expressions (10) and (12) – (17):

When the external electromotive force of AC-generator (generator voltage) has a simple harmonic view:

e_ext(t) = e_m× cos(W×t+j_0e) , (10)

where e_m=e_RMSÖ2 – generator peak voltage (e_RMS – its root mean square value voltmeter of which is indicated); W=2pf – cyclic frequency of generator voltage (f –frequency of generator); j_0e– initial phase of generator voltage,

then steady state oscillations ofpublic current in a circuit will be described by an equation:

i(t) = Im×cos(Wt+j0I);	(11)
Im=em / Z – peak current,
j0I = j0e+DF – initial phase of current.

Impedance of series oscillatory circuit

, (12)

where R – resistance of cirquit;

XC =1/ WC – capacitive reactance;	(13)
XL= WL – inductive reactance;

Phase difference (DF=j_0I – j_0e) between the current and generator voltage

DF= arctg[(X_C – X_L) / R] , (14)

CONSIDERATION: if to take into account that the initial phase of current j_0I = j_0e–DF, then DF= arctg[(X_L – X_C) / R].

The public current in a circuit i(t) determines individual voltages on circuit devices:

Voltage on resistor

uR(t) = i(t)×R = UmR×cos(Wt+j0R);	(15)
UmR= ImR – peak voltage,
j0R = j0I = j0e+DF – initial phase.

Voltage on inductor

uL(t)= –eBACK= L×di(t) / dt = UmL×cos(Wt+j0L);	(16)
UmL=ImXL – peak voltage,
j0L =j0I+p/2=j0e+DF+p/2 – initial phase.

Voltage on capacitor

uC(t)= = = UmC×cos(Wt+j0C),	(17)
UmC= ImXC – peak voltage,
j0C = j0I –p/2= j0e+DF–p/2 – initial phase.

The peak current as well as peak voltages on circuit devices are strongly depends on generator frequency. Magnitude of a current and distribution between the voltages in series RLC-circuit at three various regions of frequency spectrum are shown below.

Low frequency	Resonance ΩR	High frequency
Reactance:
XC > XL. Considerably predominate capacitive reactance.	XC = XL; Lowest resistance: Z=R	XC < XL. Considerably predominate inductive reactance.
Reactive voltage:
UmL< UmC =XC×em / Z. Predominate capacitive voltage.	UmC = UmL =Q×em;	UmC < UmL =XL×em / Z. Predominate inductive voltage.
Current:
. Low current.	Im = em /R; Heavy current.	. Low current.
Phase difference (DF=j0I – j0e) between the current and generator voltage:
DF > 0;	DF = 0;	DF < 0.
Current and voltage phasors at t=0 are represented on vector voltage diagram:

EXAMPLE OF PROBLEM SOLUTION

Example 3. Simple harmonic external EMF with the frequency 10³ Hz applied to the series oscillatory circuit of RLC-filter. The elements of filter have a nominal value: resistance 100 W, inductance 40 mH and capacity 1 mF. The voltage on capacitor varies with time by the law: u_C(t) = U_mC× cos(Ωt), with the reading of an voltmeter U_{RMS C} =20 V .

1) To rebuild the equation of changing of current in circuit, of voltage on resistor, voltage on capacitor, of voltage on inductor and EMF, applied to circuit with numerical coefficients.

2) To build the vector voltage diagram at t=0.

3) To find the values of external EMF – ε, voltages – u_R, u_C , u_L at the moment of time of t₁ = Т/8 (Т – period of oscillations). To build the voltage diagram at t₁ = Т/8.

Input data: R = 100 Ω; L = 40 mH =0,04 H; С = 1 mF = 10–6 F; uC(t) = Umc cos(Ωt); URMS C = 20 V; f= 103 Hz; t1 = T/8.
Find: 1) i(t), uR(t), uL(t), uC(t), ε(t) – ? 2) Phasors at t=0 – ? 3) phasors at t1, uR(t1), uL(t1), uC(t1), ε(t1) – ?

Solution:

1) Let’s evaluate the peak voltage on capacitor U_mC and cyclic frequency W of oscillations in circuit:

U_mC =U_{rms C} ×Ö2; [U_mC]=V; U_mC = 20 ×Ö2=28,3 V;

W=2pf; [W]=rad×Hz=rad /s; W= 6,28×10³ rad /s.

Here U_{rms C} – root mean square value of voltage on capacitor voltmeter of which is indicated; f –frequency of external EMF of generator.

From the view of the given equation u_C(t) = U_mc cos(Ωt) we can conclude then j_0C =0.

Finally we obtain the equation of oscillations of voltage on capacitor with numerical coefficients:

u_С(t)= 28,3cos(6,28×10³t) V.

From Ohm’s law for the AC voltage on capacitor with the peak voltage and initial phase there is a view:

uC(t)= UmC×cos(Wt+j0C);	(3.1)
UmC= ImXC ;
j0C = j0I –p/2.

where capacitive reactance is defined by the equation:

X_C =1 / WC ; (3.2)

From this we obtain: I_m= U_mC / X_C ; j_0I =j_0C+p/2.

Let’s check dimensionality and make the calculations:

[X_C]=1/ (F× rad /s)=W; X_C = 1 / (6,28×10³×10^–6)= 159W;

[I_m]= V / W= A; I_m = 28,3 / 159=0,178 A;

[j_0I]=rad; j_0I =0+p/2=p/2.

From Ohm’s law for the AC public current in circuit with the peak value and initial phase there is a view:

i(t) = Im×cos(Wt+j0I);	(3.3)
Im=em / Z ;
j0I = j0e+DF.

where impedance of series oscillatory circuit and phase difference between the current and generator voltage are defined by the equations:

; (3.4)

DF= arctg[(X_C – X_L) / R]. (3.5)

Let’s substitute numerical values in (3.3) and obtain the equation of current oscillations with numerical coefficients:

i(t)=0,178×cos(6,28×10³t+p/2) А.

Thus current in circuit has phase lead relative to the voltage on capacitor on p/2.

From Ohm’s law for the AC voltage on resistor with the peak voltage and initial phase there is a view:

uR(t) = UmR×cos(Wt+j0R);	(3.6)
UmR= ImR ;
j0R = j0I = j0e+DF.

Let’s check dimensionality and make the calculations:

[U_mR]= A×W= V; U_mR = 0,178×100=17,8 V;

[j_0R]=rad; j_0R =p/2.

Let’s substitute numerical values in (3.6) and obtain the equation of oscillations of voltage on resistor with numerical coefficients:

u_R(t)=17,8×cos(6,28×10³t+p/2) V.

Thus voltage on resistor has phase lead relative to the voltage on capacitor on p/2.

From Ohm’s law for the AC voltage on inductor with the peak voltage and initial phase there is a view:

uL(t)= UmL×cos(Wt+j0L);	(3.7)
UmL=ImXL ;
j0L =j0I+p/2.

where inductive reactance is defined by the equation:

X_L = WL ; (3.8)

Let’s check dimensionality and make the calculations:

[X_L]=rad /s×H=W; X_L = 6,28×10³×0,04= 251 W;

[U_mL]= A×W= V; U_mL = 0,178×251=44,7 V;

[j_0I ]=rad; j_0I =p/2+p/2=p.

Let’s substitute numerical values in (3.7) and obtain the equation of oscillations of voltage on inductor with numerical coefficients:

u_L (t)=44,7cos(6,28×10³t+p) V.

Thus voltage on resistor has phase lead relative to the voltage on capacitor on p.

From Ohm’s law for the AC external EMF of a generator there is a view:

e_ext(t) = e_m× cos(W×t+j_0e), (3.9)

where the peak external voltage and the initial phase we find from equations (3.3):

e_m = I_m× Z;(3.10)

j_0e =j_0I –DF; (3.11)

Let’s check dimensionality and make the calculations:

; ;

[e_m]= A×W= V; e_m = 0,178×136=24,1 V;

[DF]=rad; DF=arctg[(159–251)/100]=arctg[–0,93]= –0,75rad= –0,24p »–p/ 4;

[j_0e]=rad; j_0e=p/2+p/4=3p/ 4.

Let’s substitute numerical values in (9) and obtain the equation of oscillations of external voltage with numerical coefficients:

e_ext(t)=24,1cos(6,28×10³t+3p/ 4) V.

Thus external voltage has phase lead relative to the voltage on capacitor on 3p/ 4.

2) For building the vector voltage diagram at t=0 let’s write the equation of voltages in this moment of time without calculating the value of cosine:

uС(t=0)= 28,3cos(0) V; uR(t=0)=17,8×cos(p/2) V; uL (t=0)=44,7cos(p) V ; eext(t=0)=24,1cos(3p/ 4) V; i(t=0)=0,178×cos(p/2) А. DF = –p/ 4; j0e= 3p/ 4.

Figure 3.4,a – The vector voltage diagram at t=0.

In this case the argument of cosine is a phase of corresponding voltage at t=0.

For each phasor, the angle which is numerically equal to an initial phase, we lay off counterclockwise (for positive phase) relative to the axis X (dashed axis at the Fig. 3.4,a). Initial phase of external voltage j_0e=3p/ 4 and DF»–p/ 4 is shown at Fig. 3.4,a.

The magnitude of each phasor is equal to peak value of corresponding voltage. If one chooses the scale of voltage equal 10 V/сm, then the phasor length, representing the oscillation of voltage on capacitor, will equal 2,8 сm (amplitude will be U_mc = 28,3 V), this vector will be directed along the bearing axis, for the reason that the argument of cosine is equal to zero at t=0.

In much the same way we build phasors, representing the oscillation of voltage on resistor, on inductor and external voltage of generator.

At the result we obtain, that according to the second Kirhchoff’s rule, the phasor external EMF must be equal to the vector sum of the all voltage phasors (see Fig. 3.4,a):

.

3) For finding values of external EMF and voltages at the moment of time of t₁ = Т/8 we calculate the period of oscillations:

T=2p / W; [T]=rad / rad/s=s; T=2×3,14 / 6,28×10³=10^–3 s.

Let’s substitute numerical value t₁ = Т/8 =10^–3/ 8 s in the equation of oscillations. For building the vector voltage diagram at t₁= Т/ 8 we underline the equation of voltages in this moment of time without calculating the value of cosine and build phasors in the same way as Fig. 3.4,a:

uС(t1)= 28,3cos(6,28/ 8)= 28,3cos(p/ 4) ; uR(t1)=17,8×cos(p/ 4+p/2)= 17,8×cos(3p/ 4); uL (t1)=44,7cos(p/ 4+p)= 44,7cos(5p/ 4); eext(t1)=24,1cos(p/ 4+3p/ 4)= 24,1cos(p); i(t1)=0,178×cos(p/ 4+p/2)= 0,178×cos(3p/ 4); DF = –p/ 4; j0e= 3p/ 4.

Figure 3.4,b – The vector voltage diagram at t1= Т/ 8.

In this case the argument of cosine is a phase of corresponding voltage at t₁ = Т/ 8. We note, that the phase of all voltages was incremented on p/ 4 and phasors have turned counterclockwise through an angle of p/ 4 (see Fig. 3.4,b).

Finally we calculate numerical values of the voltages at t₁ = Т/8:

u_С(t₁)=28,3×0,707=20 V;

i(t₁)=0,178×0,707=0,126 А;

u_R (t₁)=17,8×0,707=12,6 V;

u_L (t₁)=44,7×(–0,707)= –31,6 V;

e_ext(t₁)=24,1×(–1)= –24,1 V.

Results:

1)	uС(t)= 28,3cos(6,28×103t) V;	i(t)=0,178×cos(6,28×103t+p/2) А;
	uR(t)=17,8×cos(6,28×103t+p/2) V;	uL (t)=44,7cos(6,28×103t+p) V;
	eext(t)=24,1cos(6,28×103t+3p/ 4) V.
2)	Phasors at t=0 see on Fig. 3.4,a.
3)	Phasors at t= Т/ 8 see on Fig. 3.4,b.
	uС(t1)=20 V; i(t1)= 0,126 А; uR(t1)=12,6 V; uL (t1)=–31,6 V; eext(t1)= –24,1 V.