Chapter 16, Problem 67P

(a)

To determine

The range ofwavenumber.

Answer to Problem 67P

  $\Delta k~\frac{1}{\Delta x}{\text{m}}^{\text{-1}}$

Explanation of Solution

Introduction:

A sound wave form originated by tuning fork of frequency centered on $f_o$ .A wave packet with N number of cycles whose length in the space is $\Delta x$ of time interval of $\Delta t$ is shown in Figure 1.

  

Figure 1:A sound waveform created by a tuning fork

Basically, the wave packet is the result of the superposition of the two or more waves, hence, the resultant of the super-position of the wave results in the spread of frequencies, $\Delta f$ .The relation between the range of frequencies $\Delta f$ and duration of the pulse $\Delta t$ (or the wave-packet) is given as below:

  $\Delta f\times \Delta t~1\dots \dots \dots \left(1\right)$

Let say, the v is the speed of the pulse. Therefore, the pulse spread in space can be written as $\Delta x=v\times \Delta t\dots \dots \left(2\right)$

The range of frequencies $\Delta f$ :implies a range in wave numbers $\Delta k$ .

As it is well known relation between the velocity, frequency and the wave number is

  $\begin{array}{l}v=\frac{\text{Δ}\omega }{\text{Δ}k}\\ \text{Δ}\omega =\text{Δ}k\times v\dots \dots \mathrm{..}\left(3\right)\end{array}$

Putting the value of v from equation 1 in equation 2,

  $\begin{array}{l}\text{Δ}\omega =\text{Δ}k\times \frac{\text{Δ}x}{\text{Δ}t}\\ \text{Δ}\omega \times \text{Δ}t=\text{Δ}k\times \text{Δ}x\dots \dots \dots \dots \dots \left(4\right)\end{array}$

Comparing equation 1 and 4, we get

  $\begin{array}{l}\Delta k\times \Delta x~1\\ \Delta k~\frac{1}{\Delta x}{\text{m}}^{\text{-1}}\end{array}$

k is the wave number.

Conclusion:

The wave number range is $\Delta k~\frac{1}{\Delta x}{\text{m}}^{\text{-1}}$

(b)

To determine

The average value of the wavelength.

Answer to Problem 67P

  $\lambda =\frac{\Delta x}{N}$

Explanation of Solution

Introduction:

A wavelength $\lambda$ is the distance travelled by the wave to complete one cycle.

N number of cycles are there in a wave packet of length $\Delta x$ .

The distance for one cycle is basically equal to the wavelength and will be

  $\frac{\Delta x}{N}\dots \dots \left(1\right)$

Hence, one can write $\lambda =\frac{\Delta x}{N}$ .

Conclusion:

The average wavelength will be $\lambda =\frac{\Delta x}{N}$

(c)

To determine

The average value of the wave number k

Answer to Problem 67P

  $k=\frac{2\pi \times N}{\Delta x}$

Explanation of Solution

Introduction:

As calculated before:

  $\Delta k\times \Delta x~1$

  $\lambda =\frac{2\pi }{k}$

N number of cycles are there in a wave packet of length $\Delta x$ .

The distance for one cycle is basically equal to the wavelength and will be

  $\frac{\Delta x}{N}\dots \dots \left(1\right)$

Hence,

  $\lambda =\frac{\Delta x}{N}----\left(2\right)$

The relation between $\lambda$ and $k$ is given below

  $\lambda =\frac{2\pi }{k}=\frac{\Delta x}{N}$

From equation 3:

  $k=\frac{2\pi \times N}{\Delta x}$

Conclusion:

The average value of the wave number kis $k=\frac{2\pi \times N}{\Delta x}$ .

(d)

To determine

Therange in angular frequencies.

Answer to Problem 67P

  $\Delta f~\frac{1}{\Delta t}$

Explanation of Solution

Introduction:

A sound wave form originated by tuning fork of frequency centered on $f_o$ . A wave packet with N number of cycles whose length in the space is $\Delta x$ of time interval of $\Delta t$ is shown in Figure 1.

  

Figure 1:A sound waveform created by a tuning fork

Basically, the wave packet is the result of the superposition of the two or more waves, hence, the resultant of the super-position of the wave results in the spread of frequencies, $\Delta f$ .

The relation between the range of frequencies $\Delta f$ and duration of the pulse $\Delta t$ (or the wave-packet) is given as below:

  $\begin{array}{l}\Delta f\times \Delta t~1\\ \Delta f~\frac{1}{\Delta t}\end{array}$

Conclusion:

The range in angular frequencies $\Delta f~\frac{1}{\Delta t}$

(e)

To determine

The frequency in terms of N and $\Delta t$ .

Answer to Problem 67P

  $f_o=\frac{\Delta t}{N}$

Explanation of Solution

Introduction:

A sound wave form originated by tuning fork of frequency centered on $f_o$ . A wave packet with N number of cycles whose length in the space is $\Delta x$ of time interval of $\Delta t$ is shown in Figure 1.

  

Figure 1:A sound waveform created by a tuning fork

Basically, the wave packet is the result of the superposition of the two or more waves, hence, the resultant of the super-position of the wave results in the spread of frequencies, $\Delta f$ .

N number of cycles are there in a wave packet of length $\Delta x$ for a duration of $\Delta t$

The time required to complete one cycle is basically equal to the frequency and will be

  $\frac{\Delta t}{N}\dots \dots \left(1\right)$

Hence, one can write $f_o=\frac{\Delta t}{N}$

Conclusion:

The frequency will be $f_o=\frac{\Delta t}{N}$

(f)

To determine

The uncertainty in N.

Answer to Problem 67P

There is an uncertainty of $\pm 1$ in the number of cycles, N.

Explanation of Solution

Introduction:

A sound wave form originated by tuning fork of frequency centered on $f_o$ .A wave packet with N number of cycles whose length in the space is $\Delta x$ of time interval of $\Delta t$ is shown in Figure 1.

  

Figure 1:A sound waveform created by a tuning fork

Basically, the wave packet is the result of the superposition of the two or more waves, hence, the resultant of the super-position of the wave results in the spread of frequencies, $\Delta f$ .

As it is clear from the Figure 1, there is a cycle which is not a complete one. Hence, the cycle can be either not present or may be present in the wave packet.

Therefore, there is an error or uncertainty of $\pm 1$ in the number of cycles, N .

Conclusion:

There is an uncertainty of $\pm 1$ in the number of cycles, N.

(g)

To determine

The uncertainty in wave number k

Answer to Problem 67P

  $k=\frac{2\pi }{\Delta x}$

Explanation of Solution

Introduction:

A sound wave form originated by tuning fork of frequency centered on $f_o$ . A wave packet with N number of cycles whose length in the space is $\Delta x$ of time interval of $\Delta t$ is shown in Figure 1.

  

Figure 1:A sound waveform created by a tuning fork

Basically, the wave packet is the result of the superposition of the two or more waves, hence, the resultant of the super-position of the wave results in the spread of frequencies, $\Delta f$ .

As calculated before:

  $\begin{array}{l}\Delta k\times \Delta x~1\\ \lambda =\frac{2\pi }{k}\end{array}$

N number of cycles are there in a wave packet of length $\Delta x$ .

The distance for one cycle is basically equal to the wavelength and will be

  $\frac{\Delta x}{N}\dots \dots \left(1\right)$

Hence, we can write $\lambda =\frac{\Delta x}{N}----\left(2\right)$

The relation between $\lambda$ and $k$ is given below;

  $\lambda =\frac{2\pi }{k}=\frac{\Delta x}{N}$

From equation 3 we obtain

  $k=\frac{2\pi \times N}{\Delta x}\dots \dots .\left(3\right)$

As calculated by the previous section that the uncertainty in N is $\pm 1$ . Therefore, uncertainty in k will be, $k=\frac{2\pi \times N}{\Delta x}$

Conclusion:

The uncertainty in k will be, $k=\frac{2\pi \times N}{\Delta x}$