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
(b) y = sin 2θ from θ = 0 to θ =
π
4
(c) y = 4et from t = 1 to t = 4
4
4
1
1 4
1 4 1/2
1  3 x3/2 
1  34
(a) Mean=
value, y
=
=
=
=
y
x
x
x
x
x
x
d
3
d
3
d


 0
4 − 0 ∫0
4 ∫0
4 ∫0
4  3 / 2  0 2 
=
( )
( )
1
1 3
1
=
43
=
2
(8) = 4
2
2
2
π /4
π /4

1
4 π /4
4  cos 2θ 
2
π 
(b) Mean value, y =
=
−
=
− cos 2   − cos 0 
y
d
θ
sin 2θ d θ =
∫0
∫


0
π
2 0
π
π
π
4

−0
4
=–
(c) Mean value,
=
y
2
π
( 0 − 1)
=
2
π
or 0.637
1 4
1 4 t
1
4 4 1
t 4
 4 e=

e − e  = 69.17
y=
dx
4 e=
dt
∫
∫
1
4 −1 1
3 1
3
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 = 2 x 2 + 5 is shown below.
1146
© 2014, John Bird
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 = 2 x 2 + 5 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
and mean value =
area under curve 56.875 56.875
= 18.96
= =
length of base
4 −1
3
4

1 4
1 4 2
1  2 x3
1  128
 2

(b) By integration, =
=
+
=
+
+ 20  −  + 5  
y
y
x
x
x
x
d
2
5
d
5

=


∫
∫
1
1
4 −1
3
3 3
 3

 1 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.
Average speed, =
v
3
1 3
1 2
1
1


4
t
+
3
d
=
t
2
t
+
3
t
=
(
)
[(18 + 9) − (0)=] ( 27 ) = 9 m/s
∫


0
0
3−0
3
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
1147
© 2014, John Bird
A sketch of y = 6 + x – x2 is shown below.
3
3
1
1 3
1
x 2 x3 
2
Mean value, =
=
+
−
=
+
− 
y
y
x
x
x
x
x
d
6
d
6

3 − −2 ∫ − 2
5 ∫ −2
5
2 3  −2
(
)
=
1
8 
(18 + 4.5 − 9 ) −  −12 + 2 +  

5
3 

=
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.
4
4

1
1  2 d3 
1 
64 
2
Mean height, h=
−
=
4
d
d
d
d
2d −  =
32 −  − (0)  = 2.67 km



∫
4−0 0
4
3  0 4 
3 

(
)
6. The velocity v of a piston moving with simple harmonic motion at any time t is given by:
v = c sin ωt, where c is a constant. Determine the mean velocity between t = 0 and t =
π /ω
π /ω
1
ω  c cos ωt 
Mean velocity, v =
−
∫0 ( c sin ωt ) d t =
π
π 
ω  0
−0
ω

c 
 π 
=
−  cos ω    − (cos 0) 
π 
 ω 

=−
1148
π
ω
c
π
(−1 − 1) =
2c
π
© 2014, John Bird
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
(a) R.m.s value =
(c) y = 25 sin θ from θ = 0 to θ = 2π
4
 1

y2 d x
 =
∫
4 − 0 0

2
1 4

3x ) d x 
(
=
∫
4 0

=
(b) R.m.s value =
 1 3 2 
 =
∫ y d x 
 3 −1 1
=
 9  64  
 =
 
 4  3 
144  12
= 6.928
=


3
 3 
1 3 2 2 
=
∫ t d x 
2 1
1 3 4 
=
 ∫1 t d x 
2

 1  243 1  
− =
 

5 
2  5
 1  242  
 

=
 2  5 
( )
 9  x3  4 
   
 4  3  0 
9 4 2 
=
 ∫0 x d x 
4

 1  t 5  3 
   
 2  5  1 
121  11
= 4.919
 =

5
 5 
(c) R.m.s value =
2π
2
 1

25sin
=
θ ) dθ 
(

∫
0
 2π − 0

=
 252

 4π
 252 2π 2

=
sin θ d θ 

∫
 2π 0

2π
sin 2θ  

−
=
θ


2  0 
 252

 2π
∫
(note that cos2 t =
(a) R.m.s value =
0

1
(1 − cos 2θ ) d θ 
2

 252

− (0) ]
[(2π − 0)=

 4π

2. Calculate the r.m.s. values of: (a) y = sin 2θ from θ = 0 to θ =
(b) y = 1 + sin t from t = 0 to t = 2π
2π
 252 
25
or 17.68

 =
2
 2 
π
4
(c) y = 3 cos 2x from x = 0 to x = π
1
(1 + cos 2t), from Chapter 44).
2


π /4
π /4
 1

sin 4θ  
 2 
 4 π /4 1

2
(1 − cos 4θ ) d θ  =
π
 ∫0
 θ −

∫0 sin 2θ d θ  =
2
4  0 
π


 π 
 −0

4

=
 2  π


  − 0  − ( 0 ) =



 π  4
1149
1 
 =

2
1
or 0.707
2
© 2014, John Bird
(b) R.m.s value =
2π
2
 1

1 + sin t ) =
dt
(

∫
 2π − 0 0

 1

 2π
=
 1

 2π
=
2π
 1  3t
sin 2t  
  − 2 cos t − =

4  0 
 2π  2
=
3
 1

= 1.225
 ( 3π )  =
2
 2π

(c) R.m.s value =
π
2
 1

2x) d x
( 3cos=

∫
0
π − 0

∫
2π
0
2π
 9

 2π
9

π
 1

 2π
∫
2π
0
1
3
 
 + 2sin t − cos 2t  d t 
2
2
 
 1

 ( 3π − 2 − 0 ) − ( 0 − 2 − 0 )  
 2π

∫
π
 sin 4 x  
x + 4 =
 

 0 
[Note that cos 2x = 2 cos 2 x − 1 and
2
0
1


1
+
2sin
t
+
1
−
cos
2
t
=
d
t
(
)

 
2

=
from which, cos 2 =
2x

∫ (1 + 2sin t + sin t ) d t 
π
0

2
=
cos
2x d x

9

π
1

∫ 2 (1 + cos 4 x ) d x 
π
0
 9
( 0 ) 
 (π + 0 ) −=
 2π

9
or 2.121
2
cos 4x = 2 cos 2 2 x − 1
1
(1 + cos 4 x ) ]
2
3. The distance p of points from the mean value of a frequency distribution are related to the
variable q by the equation p =
1
+ 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

 1 3  1

+
=
q
dq




3 − 1 ∫1  q



Standard deviation = r.m.s. value =
=
1 3 1

2
 ∫1  2 + 2 + q  dq 
 
2  q
3
−1
q 3  
 1  q
+ 2q + =
 
 
3 1 
 2  −1

 1  1
1  
 
   − + 6 − 9  −  −1 + 2 +   
3  
 
 2  2
=
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.
1150
© 2014, John Bird
10×10−3
10×10−3
1
 30

Mean value =
30sin100π t =
d t 100  −
cos100π t 
(
)
−3
∫
10 ×10 − 0 0
 100π
0
= −
100(30)
cos(100π ×10 ×10−3 ) − cos 0 

100π
= −
30
π
30
60
π
π
[cos π − cos 0] =− [ −1 − 1] =
= 19.10 A
r.m.s. value =
10×10−3
1
302 sin 2 100π t d t
=
−3
∫
0
10 ×10 − 0
(100)(30) 2 ∫
10×10−3
0
1
(1 − cos 200π t ) d t
2
1
(1 − cos 2 A)
2
since cos 2A = 1 – 2sin 2 A from which, sin 2=
A
10×10−3
 sin 200π t 
=
t − 200
π  0

=
(100)(30) 2
2
=
(100)(30) 2
10 ×10−3  =
2
(100)(30) 2 
sin 2π
−3
10 ×10 −

2
200π



 − (0 − sin 0) 


302 30
= 21.21 A
=
2
2
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,
mean value =
and
r.m.s. value =
2
π
2
× peak value =
× 340 = 216 V
π
1
1
× peak value =
× 340 = 240 V
2
2
6. Determine the form factor, correct to 3 significant figures, of a sinusoidal voltage of maximum
value 100 volts, given that form factor =
r.m.s.value
average value
For a sine wave,
1151
© 2014, John Bird
average value =
form factor =
π
× peak value=
2
π
×100 = 63.66 V
1
1
× peak value=
×100 = 70.71 V
2
2
r.m.s. value =
and
2
r.m.s.value
70.71
= 1.11
=
average value 63.66
7. A wave is defined by the equation:
v = E1 sin ωt + E3 sin 3ωt
where E1 , E3 and ω are constants.
Determine the r.m.s. value of v over the interval 0 ≤ t ≤
1
r.m.s. value =
π
−0
ω
∫
∫
∫
0
π
ω
0
π
ω
0
sin 2 ωt d t =
sin 2 3ωt d t =
∫ ( E sin ωt + E
0
1
3
sin 3ωt ) d(ωt )
2
ω ωπ 2 2
E1 sin ωt + 2 E1 E3 sin ωt sin 3ωt + E32 sin 2 3ωt ) d(ωt )
(
∫
0
π
=
π
ω
π
ω
π
ω
∫
π
ω
1 − cos 2ωt
dt =
2
0
∫
π
ω
0
π
 π
 t sin 2ωt  ω  π

 2 − 4ω  =  2ω − 0  − (0 − 0)  = 2ω


0
1 − cos 6ωt
dt =
2
π
 π
 t sin 6ωt  ω  π

 2 − 12ω  =  2ω − 0  − (0 − 0)  = 2ω


0
π
π
1
ω sin 3ωt sin ωt d t =
ω −
sin ωt sin 3ωt d t =
∫0
∫0 2 ( cos 4ωt − cos 2ωt ) d t
π
1  sin 4ωt sin 2ωt  ω
1
0
=
− 
−
=
− [ (0 − 0) − (0 − 0) ] =

2  4ω
2ω  0
2
Hence, r.m.s. value =
 E12 E32 
ω  π 2 π 2
+
=
+
E
E


π  2ω 1 2ω 3 
2 
 2
=
 E12 + E32 


2


1152
© 2014, John Bird