The essence of physics is to describe and verify the relationships among physical quantities. The relationships are often simple. For example, two quantities may be directly proportional to each other, inversely proportional to each other, or one quantity may be inversely proportional to the square of the other quantity.
Direct Relationship
Two quantities are said to be directly proportionalto one another if an increase (or decrease) of the first quantity causes an increase (or decrease) of the second quan- tity by the same factor. If yis directly proportional to x, the direct proportionality is written as y x. The ratio y x is a constant, say,k. That is, For example, the ratio of the circumference Cto the diameter dof a circle is always 3.14, known as (pi). Therefore, the circumference of a circle is directly propor- tional to its diameter as where is the constant of proportionality.
Another simple example of direct proportionality is the stretching or compression of an ordinary helical spring (discussed in Section 5.4). The spring has a certain
p C5 pd,
p
y1 x15 y2
x25k.
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EXAMPLE 0.7 Solving two symbolic equations in two unknowns
Use the equations v5v01atand x5v0t112at2to obtain an equation for xthat does not contain a.
S O L U T I O N
S E T U P A N D S O LV E We solve the first equation for a:
We substitute this expression into the second equation:
51 2v0t11
2vt5 ¢v01v 2 ≤t.
x5v0t11 2¢v2v0
t ≤t25v0t11 2vt21
2v0t a5v2v0
t .
R E F L E C T When you solve a physics problem, it’s often best to work with symbols for all but the final step of the problem. Once you’ve arrived at the final equation, you can plug in numerical values and solve for an answer.
EXAMPLE 0.6 Solving two equations in two unknowns
Solve the following pair of equations for xand y:
3x25y5 29 x14y514
S O L U T I O N
S E T U P A N D S O LV E The first equation gives
Substituting this for xin the second equation yields, successively, and
Thus, Then We can
verify that satisfies both equations.
An alternative approach is to multiply the first equation by
yielding Adding this to the second
equation gives, successively,23x212y5 242.3x25y1 123x2 1 1212y2 5 23,
y53 x52,
1421252.
x51424y5
25121753.
y5 42212y25y5 29, 217y5251.
311424y225y5 29,
x51424y. and which agrees with our
previous result.
R E F L E C T As shown by the alternative approach, simultaneous equations can be solved in more than one way. The basic meth- ods we describe are easy to keep straight; other methods may be quicker, but may require more insight or forethought. Use the method you’re comfortable with.
y53, 217y5 251,
291 12422,
length at rest, which increases when it is suspended vertically with weights attached to the bottom. If the amount of stretch is not too long, the amount of forceF(measured in pounds or newtons) on the spring and the amount of stretch xare directly proportional to each other (Figure 0.1). Thus, where kis the constant of proportionality.
In general, in a direct proportion, . Multiplying both sides by bd, we find
Graph of Direct Proportionality Relationship
Wheny is directly proportional to x, and the graph of yversus xis a straight line passing through the origin as shown in Figure 0.2. In the graph, the change of the quantity xis labeled as x(which is often called “run”) and the corresponding change in yis labeled as y(which is often called “rise”).
Here,
where (x1, y1) and (x2, y2) are coordinates of the two points on the line. The constant of proportionality between yand xis also k. Thus,
The steepness of the line is measured by the ratio y xand is called the slopeof the line. Thus,
The slope of a line can be positive,negative,zero,orundefined, as shown in Figure 0.3 (a–d).
Slope5 Dy Dx5k
/D D Dy5kDx
D D
Dy5y22y1, Dx5x22x1
DD y5kx, b⁄d#a
b⁄ 5bd⁄ # c
d⁄ or a#d5b#c
a b5 c
d
F5kx
0.4 Direct, Inverse, and Inverse-Square Relationships 0-7
x5 0.1 m
F5 2.0 N
䊱FIGURE 0.1
x Dy
(x2,y2)
(x1,y1) Dx y
䊱FIGURE 0.2
䊱FIGURE 0.3 Slope positive
x y
(a)
Slope negative
x y
(b)
Slope zero
x y
(c)
Slope undefined
x y
(d)
Note that: slope positivemeans y increases as x increases; slope negative meansy decreases as x increases; slope zeromeansydoes not change—that is, the line is parallel to the x-axis; slope undefinedmeansxdoes not change—that is, the line is parallel to the y-axis.
Inverse Proportion
When one quantity increases and the other quantity decreases in such a way that their product stays the same, they are said to be in inverse proportion. In inverse proportion, when one quantity approaches zero, the other quantity becomes extremely large, so that the product remains the same. For example, the product of the pressure and volume of an ideal gas remains constant if the temperature of the gas is maintained constant, as you will find in Section 15.2. Mathematically, if yis in inverse proportion to x,y 1x. This gives y k x, or xy k, where kis the constant of proportionality. That is, when xchanges from x1to x2,ychanges from y1
to y2so that x1 y1 x2 y2 k. This type of behavior is illustrated in Figure 0.4 (for an arbitrary choice of k5 5100).5
/ 5
/ 5
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EXAMPLE 0.8 Solving for a quantity in direct proportion
If yis directly proportional to x, and when what x52 y58, is ywhen x510?
S O L U T I O N
S E T U P A N D S O LV E Since yis directly proportional to x,
Substituting the values we get
Multiplying both sides by 2 and 10 to get rid of the fractions gives 2y510#8580.
y 1058
2. y1
x15y2 x25k.
Divide by 2 to isolate y:
R E F L E C T Although this is a simple problem, this gives you the strategy for how to solve problems in direct proportion. Note that xhas increased by a factor of 5, so ymust also increase by the same factor.
y580 2 540.
EXAMPLE 0.9 Solving for the stiffness constant of a spring
A spring is suspended vertically from a fixed support. When a weight of 2.0 newtons is attached to the bot- tom of the spring, the spring stretches by 0.1 m. (The newton, abbreviated N, is the SI unit for force; the hanging weight exerts a downward force on the spring.) Determine the spring constant of the spring.
S O L U T I O N
S E T U P A N D S O LV E We have force, and stretch, The sketch of the problem is shown in Figure 0.1. The applied weight and the amount of stretch are related by a direct propor- tion, as expressed by F kx. Using this equation we can solve for the stiffness constant:
5 x50.10 m.
F52. 0 N
R E F L E C TThe stiffness constant in this equation is the constant of proportionality in the direct proportion between the force and the amount of stretch. Its unit is the ratio of the units of Fand x.
k5F
x 5 2.0 N
0.10 m 520 N/m
x y
25 20 15 10 5
00 5 10 15 20 25
䊱FIGURE 0.4
EXAMPLE 0.10 Solving for a quantity (volume of an ideal gas) in inverse proportion According to Boyle’s law, if the temperature of an ideal gas is kept constant, its pressure,P, is inversely proportionalto its volume,V. A cylindrical flask is fitted with an airtight piston and contains an ideal gas.
Initially, the pressure of the inside gas is 11 104pascals (Pa) and the volume of the gas is 8.0 10–3m3. Assuming that the system is always at the temperature of 330 kelvins (K), determine the volume of the gas when its pressure increases to 243104Pa.
3 3
Inverse Square Proportion
Inverse square dependence is common in the laws of nature. For example, the force of gravity due to a body decreases as the inverse square of the distance from the body, as expressed by Newton’s law of gravitation in Section 6.3. Similarly, the electrostatic force due to a point electric charge decreases as the square of the distance from the charge, as expressed by Coulomb’s law in Section 17.4. The intensity of sound and of light also decreases as the inverse square of the distance from a point source, as you will find in Section 12.10. (Intensity in these cases is a measure of the power of the sound or light per unit area.) Mathematically, if y varies inversely with the square of x, then
y or y or y k,
where kis the constant of proportionality. That is, when xchanges from x1to x2,y changes from y1to y2so that x1
2y1 x2
2y2 k. Or
This relationship is illustrated in Figure 0.5 (for an arbitrary choice of k5100).
y1 y25 x22
x21
5 5
5 x2 k
x2 1 5
x2
~
0.4 Direct, Inverse, and Inverse-Square Relationships 0-9
x y
25 20 15 10 5
00 5 10 15 20 25
䊱FIGURE 0.5 S O L U T I O N
S E T U P A N D S O LV E Since the pressure Pis inversely propor- tional to the volume V according to Boyle’s law, the product PV remains constant.
That is,
P1V1 P2V2. We divide by P2to solve for V2.V2
In this problem,P1 11 104Pa,V1 8.0 10–3m3, and
P25243104Pa. 5 3 5 3
P1V1 P2 . 5 5
Substituting the values of P1,V1, and P2, we solve for V2.
V2 10–3m3
3.7 10–3m3
R E F L E C T Since the pressure increased, the final volume has decreased, as expected in an inverse proportion. Note that the pascal (Pa) and K kelvin (K) are the SI units for pressure and temperature, respectively./
3 5
1138.03 5 24 1113104 Pa2 3 18.031023 m32
243104 Pa 5
EXAMPLE 0.11 Solving for a quantity (sound intensity) that varies as the inverse square
A small source of sound emits sound equally in all directions. The intensity of emitted sound at a point is given by the equation I k r2, where k is a constant of proportionality and ris the distance of the point from the source. If the sound intensity is 0.05 watt meter2(W m2) at a distance 0.5 m from the source, find the sound intensity at a distance of 20 m from the source.5 / / /
S O L U T I O N
S E T U P A N D S O LV E The intensity of sound is given by the equation I k r2. We apply this equation to solve the problem.
However, we do not need to know the value of the constant k. In this problem, we have an initial distance, r1 0.5 m, sound intensity, I1 0.05 W m2, and final distance,r2 20 m, where intensity I2is to be determined.
From the inverse square relationship, we have (where the vari- ables xand yhave been replaced by r and Iin the equation)
I1 I25r22
r21
/ 5
5 5
5 /
We take the reciprocal of both sides and multiply by I1to solve for I2.
I2 I1 (0.05 W m2)
R E F L E C T As the distance increases, the intensity decreases. Note that because the intensity decreases as the square of the distance, the result for intensity is less than it would have been for the case of a simple inverse proportion. Thus, the result makes sense.
53.131025 W/m2
10.5 m22 120 m22
/
r215 r22 5