The student should draw the two graphs viz.
a graph of P against v2 and a graph of
P against v-2 using the values in the table below.
| P x 103 Nm-2 |
2.0 |
1.5 |
1.2 |
0.7 |
0.3 |
| v ms-1 |
1.0 |
1.4 |
1.6 |
1.9 |
2.1 |
| v2m2s-2 |
1.00 |
1.96 |
2.56 |
3.61 |
4.41 |
| (1/v2)m-2s2 |
1.00 |
0.50 |
0.39 |
0.28 |
0.23 |
Results:
A plot of the experimental values of P against v-2 yields
a curve contrary to what is predicted by equation (3). Therefore
equation (3) is wrong.
A plot of P against v2 yields a straight line
with a negative gradient as predicted by equation (2). Therefore
P + Brv2 = Y is the correct equation.
Gradient of the graph = - 2.48 x 103 /5.0 = -
Br
Given r = 103 kgm-3
B = (0.5 x 103)/103 = 0.5
when v2 = 0, P = Y, the intercept on the P - axis.
Y = 2.48 x103 Nm-2
Deducing relationships between variables
of a physical system:
The first step is to specify the various factors involved.
The unknown functional relationship is then determined by
the method of dimensional analysis, except for the dimensionless
constants.
Illustrations of the application of method of dimensional
analysis.
(a) It is required to obtain an expression for the period
T, of oscillation of a simple pendulum.
We can reasonably assume that the period T will depend on:
the length L of the string, the mass m of the bob and the
acceleration g due to gravity.
The relation between T and these variables can be expressed
as
T = kLx mygz where x, y
and z are numbers and k is a dimensionless constant.
This assumes that each quantity enters a definite power in
the function.
Balancing the dimensions of both sides of the function.
T = [L]x [M]y [LT-2 ]z
M0L0 T-1 = Lx+z
My T-2z
Hence, equating the indices for M,L,T separately on both
sides of the equation.
M: 0 = y ; L: 0 = x +z ; T : -1 = -2z
Solving y = 0 ; x = - ½ ; z = - ½
y = 0 means that the period T does not depend on the mass
m as earlier assumed.
T = kL½ g-½ = k Ö(L/g)
Note: k can not be obtained by dimensional analysis since
it is dimensionless. The value of k can be found by experiment
or from a detailed mechanical solution of the problem.
Experimental determination of the value of k.
Using T = k Ö(L/g)
T2 = ( k2/g )L of the form y = mx.
A graph of T2 against L is linear with a slope
s = k2/g
k = (sg)½
In the experiment, the length L of the simple pendulum is
varied and the time for (say) 10 oscillations is obtained
in each case, and the period T calculated. A graph of T2
against L is plotted and k is obtained.
The following set of data is obtained for different lengths
L of a simple pendulum.
| length L(m) |
0.20 |
0.40 |
0.60 |
0.80 |
1.00 |
| Period T (s) |
1.00 |
1.34 |
1.61 |
1.84 |
2.03 |
|
Plot a graph of T2 against L and use
it to determine the value of k. ( use g = 9.8m/s2
)
|
