Section 1-4
Section 1-4 Standing Waves in a Clamped String
where λ depends on the elasticity of the medium. The combination of partial derivatives on the left-hand side of Eq. (1-21) is called the Laplacian, and is often given the shorthand symbol ∇2 (del squared). This would give for Eq. (1-21) ∇2ψ(x, y, z) = −(2π/λ)2ψ(x, y, z)
(1-22)
1-4 Standing Waves in a Clamped String We now demonstrate how Eq. (1-20) can be used to predict the nature of standing waves in a string. Suppose that the string is clamped at x = 0 and L. This means that the string cannot oscillate at these points. Mathematically this means that ψ(0) = ψ(L) = 0
(1-23)
Conditions such as these are called boundary conditions. Our question is, “What functions ψ satisfy Eq. (1-20) and also Eq. (1-23)?” We begin by trying to find the most general equation that can satisfy Eq. (1-20). We have already seen that A sin(2π x/λ) is a solution, but it is easy to show that A cos(2π x/λ) is also a solution. More general than either of these is the linear combination3 ψ(x) = A sin(2πx/λ) + B cos(2πx/λ)
(1-24)
By varying A and B, we can get different functions ψ.
There are two remarks to be made at this point. First, some readers will have noticed that other functions exist that satisfy Eq. (1-20). These are A exp(2π ix/λ)
√
and A exp(−2πix/λ), where i = −1. The reason we have not included these in the general function (1-24) is that these two exponential functions are mathematically equivalent to the trigonometric functions. The relationship is exp(±ikx) = cos(kx) ± i sin(kx).
(1-25)
This means that any trigonometric function may be expressed in terms of such exponentials and vice versa. Hence, the set of trigonometric functions and the set of exponentials is redundant, and no additional flexibility would result by including exponentials in Eq. (1-24) (see Problem 1-1). The two sets of functions are linearly dependent.4
The second remark is that for a given A and B the function described by Eq. (1-24)
is a single sinusoidal wave with wavelength λ. By altering the ratio of A to B, we cause the wave to shift to the left or right with respect to the origin. If A = 1 and B = 0, the wave has a node at x = 0. If A = 0 and B = 1, the wave has an antinode at x = 0.
We now proceed by letting the boundary conditions determine the constants A and B.
The condition at x = 0 gives ψ(0) = A sin(0) + B cos(0) = 0
(1-26)
3Given functions f1, f2, f3 . . . . A linear combination of these functions is c1f1 + c2f2 + c3f3 + · · · , where c1, c2, c3, . . . are numbers (which need not be real).
4If one member of a set of functions (f1, f2, f3, . . . ) can be expressed as a linear combination of the remaining functions (i.e., if f1 = c2f2 + c3f3 + · · · ), the set of functions is said to be linearly dependent. Otherwise, they are linearly independent.
Chapter 1 Classical Waves and the Time-Independent Schr ¨odinger Wave Equation
However, since sin(0) = 0 and cos(0) = 1, this gives B = 0
(1-27)
Therefore, our first boundary condition forces B to be zero and leaves us with ψ(x) = A sin(2πx/λ)
(1-28)
Our second boundary condition, at x = L, gives ψ(L) = A sin(2πL/λ) = 0
(1-29)
One solution is provided by setting A equal to zero. This gives ψ = 0, which corresponds to no wave at all in the string. This is possible, but not very interesting. The other possibility is for 2π L/λ to be equal to 0, ±π, ±2π, . . . , ±nπ, . . . since the sine function vanishes then. This gives the relation 2π L/λ = nπ, n = 0, ±1, ±2, . . .
(1-30)
or λ = 2L/n, n = 0, ±1, ±2, . . .
(1-31)
Substituting this expression for λ into Eq. (1-28) gives ψ(x) = A sin(nπx/L), n = 0, ±1, ±2, . . .
(1-32)
Some of these solutions are sketched in Fig. 1-5. The solution for n = 0 is again the uninteresting ψ = 0 case. Furthermore, since sin(−x) equals −sin(x), it is clear that the set of functions produced by positive integers n is not physically different from the set produced by negative n, so we may arbitrarily restrict our attention to solutions with positive n. (The two sets are linearly dependent.) The constant A is still undetermined.
It affects the amplitude of the wave. To determine A would require knowing how much energy is stored in the wave, that is, how hard the string was plucked.
It is evident that there are an infinite number of acceptable solutions, each one corresponding to a different number of half-waves fitting between 0 and L. But an even larger infinity of waves has been excluded by the boundary conditions—namely, all waves having wavelengths not divisible into 2L an integral number of times. The result
Figure 1-5
Solutions for the time-independent wave equation in one dimension with boundary conditions ψ(0) = ψ(L) = 0.
where λ depends on the elasticity of the medium. The combination of partial derivatives on the left-hand side of Eq. (1-21) is called the Laplacian, and is often given the shorthand symbol ∇2 (del squared). This would give for Eq. (1-21) ∇2ψ(x, y, z) = −(2π/λ)2ψ(x, y, z)
(1-22)
1-4 Standing Waves in a Clamped String We now demonstrate how Eq. (1-20) can be used to predict the nature of standing waves in a string. Suppose that the string is clamped at x = 0 and L. This means that the string cannot oscillate at these points. Mathematically this means that ψ(0) = ψ(L) = 0
(1-23)
Conditions such as these are called boundary conditions. Our question is, “What functions ψ satisfy Eq. (1-20) and also Eq. (1-23)?” We begin by trying to find the most general equation that can satisfy Eq. (1-20). We have already seen that A sin(2π x/λ) is a solution, but it is easy to show that A cos(2π x/λ) is also a solution. More general than either of these is the linear combination3 ψ(x) = A sin(2πx/λ) + B cos(2πx/λ)
(1-24)
By varying A and B, we can get different functions ψ.
There are two remarks to be made at this point. First, some readers will have noticed that other functions exist that satisfy Eq. (1-20). These are A exp(2π ix/λ)
√
and A exp(−2πix/λ), where i = −1. The reason we have not included these in the general function (1-24) is that these two exponential functions are mathematically equivalent to the trigonometric functions. The relationship is exp(±ikx) = cos(kx) ± i sin(kx).
(1-25)
This means that any trigonometric function may be expressed in terms of such exponentials and vice versa. Hence, the set of trigonometric functions and the set of exponentials is redundant, and no additional flexibility would result by including exponentials in Eq. (1-24) (see Problem 1-1). The two sets of functions are linearly dependent.4
The second remark is that for a given A and B the function described by Eq. (1-24)
is a single sinusoidal wave with wavelength λ. By altering the ratio of A to B, we cause the wave to shift to the left or right with respect to the origin. If A = 1 and B = 0, the wave has a node at x = 0. If A = 0 and B = 1, the wave has an antinode at x = 0.
We now proceed by letting the boundary conditions determine the constants A and B.
The condition at x = 0 gives ψ(0) = A sin(0) + B cos(0) = 0
(1-26)
3Given functions f1, f2, f3 . . . . A linear combination of these functions is c1f1 + c2f2 + c3f3 + · · · , where c1, c2, c3, . . . are numbers (which need not be real).
4If one member of a set of functions (f1, f2, f3, . . . ) can be expressed as a linear combination of the remaining functions (i.e., if f1 = c2f2 + c3f3 + · · · ), the set of functions is said to be linearly dependent. Otherwise, they are linearly independent.
Chapter 1 Classical Waves and the Time-Independent Schr ¨odinger Wave Equation
However, since sin(0) = 0 and cos(0) = 1, this gives B = 0
(1-27)
Therefore, our first boundary condition forces B to be zero and leaves us with ψ(x) = A sin(2πx/λ)
(1-28)
Our second boundary condition, at x = L, gives ψ(L) = A sin(2πL/λ) = 0
(1-29)
One solution is provided by setting A equal to zero. This gives ψ = 0, which corresponds to no wave at all in the string. This is possible, but not very interesting. The other possibility is for 2π L/λ to be equal to 0, ±π, ±2π, . . . , ±nπ, . . . since the sine function vanishes then. This gives the relation 2π L/λ = nπ, n = 0, ±1, ±2, . . .
(1-30)
or λ = 2L/n, n = 0, ±1, ±2, . . .
(1-31)
Substituting this expression for λ into Eq. (1-28) gives ψ(x) = A sin(nπx/L), n = 0, ±1, ±2, . . .
(1-32)
Some of these solutions are sketched in Fig. 1-5. The solution for n = 0 is again the uninteresting ψ = 0 case. Furthermore, since sin(−x) equals −sin(x), it is clear that the set of functions produced by positive integers n is not physically different from the set produced by negative n, so we may arbitrarily restrict our attention to solutions with positive n. (The two sets are linearly dependent.) The constant A is still undetermined.
It affects the amplitude of the wave. To determine A would require knowing how much energy is stored in the wave, that is, how hard the string was plucked.
It is evident that there are an infinite number of acceptable solutions, each one corresponding to a different number of half-waves fitting between 0 and L. But an even larger infinity of waves has been excluded by the boundary conditions—namely, all waves having wavelengths not divisible into 2L an integral number of times. The result
Figure 1-5
Solutions for the time-independent wave equation in one dimension with boundary conditions ψ(0) = ψ(L) = 0.