Chapter 73 Mean and Root Mean Square Values

CHAPTER 73 MEAN AND ROOT MEAN SQUARE VALUES

EXERCISE 286 Page 778

1. Determine the mean value of (a) y = 3 x from x = 0 to x = 4

π (b) y = sin 2θ from θ = 0 to θ = (c) y = 4et from t = 1 to t = 4 4

3/2 4 44 4 4 1 1 11/2 13x 13 (a) Mean value, y= ∫∫ yxd3d3d= xx= ∫ x x= = x − 00 0 0 4 0 4 4 4 3/20 2

1 11 = 433=( 2) = (8) = 4 2( ) 22

π /4 1 ππ/4 4 /4 4 cos 2θπ  2  (b) Mean value, yy=dθ = sin 2 θθ d =−=−−cos 2 cos 0 π ∫∫00ππ24 π − 0  0  4 2 2 = – (01− ) = or 0.637 π π

144 1 144 (c) Mean value, y= yxd= 4ett d t = 4e =  e41 − e  = 69.17 41− ∫∫11 3 31 3  

2. Calculate the mean value of y = 2x2 + 5 in the range x = 1 to x = 4 by (a) the mid-ordinate rule and

(b) integration.

(a) A sketch of y = 25x2 + is shown below.

Using 6 intervals each of width 0.5 gives mid-ordinates at x = 1.25, 1.75, 2.25, 2.75, 3.25 and

3.75, as shown in the diagram

x 1.25 1.75 2.25 2.75 3.25 3.75 y = 25x2 + 8.125 11.125 15.125 20.125 26.125 33.125

Area under curve between x = 1 and x = 4 using the mid-ordinate rule with 6 intervals

≈ (width of interval)(sum of mid-ordinates)

≈ (0.5)( 8.125 + 11.125 + 15.125 + 20.125 + 26.125 + 33.125)

≈ (0.5)(113.75) = 56.875

area under curve 56.875 56.875 and mean value = = = = 18.96 length of base 4− 1 3

3 4 44 1 1 2 1 2x 1 128  2  (b) By integration, y= yxd = 2 x + 5 d x = +5 x  =  +20  −+ 5  − ∫∫11 41 3 3 31 3 3  3 

1 = (57) = 19 3

This is the precise answer and could be obtained by an approximate method as long as sufficient intervals were taken

3. The speed v of a vehicle is given by: v = (4t + 3) m/s, where t is the time in seconds. Determine

the average value of the speed from t = 0 to t = 3 s.

13 113 1 Average speed, v=(4 t + 3) d t = 2 tt2 + 3 =[ (18 +− 9) (0)] =( 27) = 9 m/s 30− ∫0 30 3 3

4. Find the mean value of the curve y = 6 + x – x2 which lies above the x-axis by using an

approximate method. Check the result using integration.

6 + x – x2 = (3 – x)(2 + x) and when y = 0 (i.e. the x-axis), then (3 – x)(2 + x) = 0 from which, x = 3 and x = –2 Hence the curve y = 6 + x – x2 cuts the x-axis at x = –2 and at x = 3 and at x = 0, y = 6

A sketch of y = 6 + x – x2 is shown below.

233 33 112 1xx Mean value, y= yxd=( 6 +− xx) d6 x = x + − −− ∫∫−−22 32 5 5 23−2

18 = (18+ 4.5 − 9) −− 12 + 2 + 53

1 = (13.5) −−( 7.3333) 5

= 4.17

5. The vertical height h km of a missile varies with the horizontal distance d km, and is given by

h = 4d – d2. Determine the mean height of the missile from d = 0 to d = 4 km.

3 4 4 1 221 d 1 64 Mean height, h=(4 dd −) d d = 2 d −  =32 −−  (0) = 2.67 km − ∫0 40 4 30 4 3

6. The velocity v of a piston moving with simple harmonic motion at any time t is given by:

π v = c sin ωt, where c is a constant. Determine the mean velocity between t = 0 and t = ω

πω/ 1 πω/ ωωctcos  cπ Mean velocity, v=( csinωω tt) d =−=−cos − (cos0) π ∫0 πω π ω − 0  0  ω c 2c = −( −− 1 1) = π π

EXERCISE 287 Page 779

1. Determine the r.m.s. values of: (a) y = 3x from x = 0 to x = 4

(b) y = t2 from t = 1 to t = 3 (c) y = 25 sin θ from θ = 0 to θ = 2π

3 4 44 4 122  12  99 x (a) R.m.s value = yxd = ( 3d x) x = xx d =   40− ∫∫00 4 4 ∫ 0  43 0

9 64  144  12 =  =  = = 6.928 43  3  3

5 3 332 3 122  1  11 4 t (b) R.m.s value = yxd = ( t) dd x = tx =   31− ∫∫11 2 2 ∫ 1  25 1

1 243 1   1  242    121 11 = −=     = = = 4.919 25 5   25     5 5

22 1 2π2 25 22 ππ25 1   ∫(25sinθ) d θ= ∫∫sin2 θθ d = (1− cos 2θ) d θ 20π− 02 ππ 002 2 

2π 25222 sin 2θ  25 25 25 =  θπ− =[(2 −− 0) (0)] = = or 17.68 ππ 420  4 22

π 2. Calculate the r.m.s. values of: (a) y = sin 2θ from θ = 0 to θ = 4 (b) y = 1 + sin t from t = 0 to t = 2π (c) y = 3 cos 2x from x = 0 to x = π

1 (note that cos2 t = (1 + cos 2t), from Chapter 44). 2

 π /4 1 ππ/4 4/4 1 2 sin 4θ  (a) R.m.s value =  sin2 2θθ d =(1 −=− cos 4θ) d θ θ  π ∫∫00ππ24 − 0   0 4

1122ππ2   (b) R.m.s value =  (1+ sint) d t =(1 ++ 2sint sin2 tt) d  20ππ− ∫∫002 

122ππ 1 13 1  = ∫∫1++− 2sint( 1 cos2 tt) d  = +−2sint cos 2 tt d  2ππ00 2 22 2 

2π 1 3tt sin 2 1  =  −2cost − =(3π −− 20) −( 020 −−)  ππ 22 4 0 2 

13 = (3π ) = = 1.225 22π

1π2 9 ππ 91  (c) R.m.s value =  (3cos 2x) d x= cos2 2xx d = (1+ cos 4x) d x π− 02∫0 ππ ∫∫ 00  

π 9 sin 4x  9 9 =  x + =(π +−00) ( )  = or 2.121 ππ  24 0 2 2

[Note that cos 2x = 2 cos2 x − 1 and cos 4x = 2 cos2 2x − 1

1 from which, cos2 2xx=( 1 + cos 4 ) ] 2

3. The distance p of points from the mean value of a frequency distribution are related to the 1 variable q by the equation p = + q. Determine the standard deviation (i.e. the r.m.s. value), q correct to 3 significant figures, for values from q = 1 to q = 3

2 33 1 1 11 2  Standard deviation = r.m.s. value =  +q dq =  ++2 q  dq − ∫∫112 31 qq2  

3 1qq−1311 1  = +2q +  =  − +69 −  −−+ 12 +  21− 3 2 2 3  1 

1 = (13.3333) = 2.58 2

4. A current, i = 30 sin 100πt amperes is applied across an electric circuit. Determine its mean and

r.m.s. values, each correct to 4 significant figures, over the range t = 0 to t = 10 ms.

10× 10−3 1 10× 10−3 30 Mean value = (30sin100ππtt) d= 100 − cos100 t −3 ∫ 0  10×− 10 0 100π 0

100(30) = − cos(100π ×× 10 10−3 ) − cos0 100π 

30 30 60 = −[cosπ − cos 0] =−[ −− 1 1] = π ππ

= 19.10 A

1110×× 10−−3310 10 r.m.s. value = 3022 sin 100ππtt d= (100)(30)2 (1− cos 200t )d t 10×− 10−3 0 ∫∫002

1 since cos 2A = 1 – 2sin2 A from which, sin2 AA= (1 − cos 2 ) 2

−3 2210× 10 (100)(30) sin 200ππt (100)(30)  −3 sin 2  = t − = 10× 10 − −−(0 sin 0) 2 200ππ0 2  200 

22 (100)(30) −3 30 30 = 10×= 10 = = 21.21 A 222

5. A sinusoidal voltage has a peak value of 340 V. Calculate its mean and r.m.s. values, correct to 3

significant figures.

For a sine wave,

22 mean value = ×=×peak value 340 = 216 V ππ

11 and r.m.s. value = ×=×peak value 340 = 240 V 22

6. Determine the form factor, correct to 3 significant figures, of a sinusoidal voltage of maximum

r.m.s.value value 100 volts, given that form factor = average value

For a sine wave,

22 average value = ××peak value= 100 = 63.66 V ππ

11 r.m.s. value = ××peak value= 100 = 70.71 V 22

r.m.s.value 70.71 and form factor = = = 1.11 average value 63.66

7. A wave is defined by the equation:

v = E13sinωω tE+ sin 3 t

where E1 , E3 and ω are constants. π Determine the r.m.s. value of v over the interval 0 ≤≤t ω

π 1 2 r.m.s. value = ω (Esinω tE+ sin 3 ωω t) d( t ) π ∫ 0 13 − 0 ω

π ω ω 22 22 = (E1sinω t++ 2 EE13 sin ω t sin 3 ω t E3 sin 3 ωω t) d( t ) π ∫ 0

π ππ 1− cos 2ωttsin 2 ωπ t ω   π ωωsin2 ωtt d = dt= − = −0 −− (0 0) = ∫∫00  2 24ωω 0  2  2 ω

π ππ 1− cos 6ωttsin 6 ωπ t ω   π ωωsin2 3ωtt d = dt= − = −0 −− (0 0) = ∫∫00  2 2 12ωω 0  2  2 ω

π ππ1 ωsinωt sin 3 ω tt d= ωω sin 3 ωω t sin tt d =−−( cos 4ωt cos 2 ω t) d t ∫∫∫0 002 π 1 sin 4ωωtt sin 2ω 1 =− − =−[(0 −−− 0) (0 0)] = 0 24ωω 20 2 ωπ π E 2 E 2 22+=+1 3 Hence, r.m.s. value = EE13 πω2 2 ω  22

EE22+ = 13 2