Chapter 3, Problem 7P

(a)

To determine

The number of modes with wavelength between $2.0\text{cm}$ to $2.1\text{cm}$.

Answer to Problem 7P

The number of modes with wavelength between $2.0\text{cm}$ to $2.1\text{cm}$ is $10$.

Explanation of Solution

Write the expression for the number of modes of vibration of waves which is confined by a cavity that is related to wavelength and length of the string is,

  $n\left(\frac{\lambda }{2}\right)=L$        (I)

Here, $n$ is the length of the string, and $\lambda$ is the wavelength.

Re-write the expression (I),

  $n=\frac{2L}{\lambda }$        (II)

Write the number of modes with wavelength between $2.0\text{cm}$ to $2.1\text{cm}$ is,

  $\begin{array}{c}dn=n_1-n_2\\ =\frac{2L}{{\lambda }_1}-\frac{2L}{{\lambda }_2}\\ =2L\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)\end{array}$        (III)

Here, $n_2$ is the number of modes related to $2.1\text{cm}$ wavelength and $n_1$ is the number of modes related to $2.1\text{cm}$ wavelength.

Conclusion:

Substitute $2.1\times {10}^{-2}\text{m}$ for ${\lambda }_2$, $2\times {10}^{-2}\text{m}$ for ${\lambda }_1$ and $2.0\text{m}$ for $L$ in expression (III),

  $\begin{array}{c}dn=2\left(2.0\text{m}\right)\left(\frac{1}{2\times {10}^{-2}\text{m}}-\frac{1}{2.1\times {10}^{-2}\text{m}}\right)\\ =10\end{array}$

Thus the number of modes with wavelength between $2.0\text{cm}$ to $2.1\text{cm}$ is $10$.

(b)

To determine

The number of modes per unit wavelength per unit length.

Answer to Problem 7P

The number of modes per unit wavelength per unit length is $0.5{\text{cm}}^{\text{-2}}$.

Explanation of Solution

Write the expression the number of modes per unit wavelength per unit length,

  $\frac{\Delta n}{L\Delta \lambda }=\frac{\Delta n}{L\left({\lambda }_2-{\lambda }_1\right)}$        (I)

Here, $n$ is the length of the string, and $L$ is the length.

Conclusion:

Substitute $10$ for $\Delta n$, $2.1\times {10}^{-2}\text{m}$ for ${\lambda }_2$, $2\times {10}^{-2}\text{m}$ for ${\lambda }_1$, and $2.0\text{m}$ for $L$ in expression (I)

  $\begin{array}{c}\frac{\Delta n}{L\Delta \lambda }=\frac{10}{\left(2.0\text{m}\right)\left(2.1\times {10}^{-2}\text{m}-\left(2\times {10}^{-2}\text{m}\right)\right)}\\ =0.5{\text{cm}}^{\text{-2}}\end{array}$

The number of modes per unit wavelength per unit length is $0.5{\text{cm}}^{\text{-2}}$.

(c)

To determine

The expression obtained the same value as found in sub part (a).

Answer to Problem 7P

The expression obtained the same value as found in sub part (a) that is the number of modes with wavelength between $2.0\text{cm}$ to $2.1\text{cm}$ is $10$.

Explanation of Solution

Write the expression the number of modes is,

  $n=\frac{2L}{\lambda }$        (I)

Here, $n$ is the length of the string, and $L$ is the length.

Re-write the expression (I) by differentiating the term, for the number of modes per unit wavelength per unit length of the string,

  $\begin{array}{c}dn=2L\left(\frac{-1}{{\lambda }^2}d\lambda \right)\\ \frac{dn}{d\lambda }=2L\left(\frac{-1}{{\lambda }^2}\right)\\ \left|\frac{dn}{d\lambda }\right|=\left|\left(\frac{-2L}{{\lambda }^2}\right)\right|\\ \frac{1}{L}\left|\frac{dn}{d\lambda }\right|=\left|\left(\frac{2}{{\lambda }^2}\right)\right|\end{array}$        (II)

Re-write the expression (II) for number of modes,

  $\left|dn\right|=\left|\left(\frac{2L}{{\lambda }^2}\right)\right|\left|d\lambda \right|$        (III)

Conclusion:

Substitute $0.1\times {10}^{-2}\text{m}$ for $\left|d\lambda \right|$, $2\times {10}^{-2}\text{m}$ for $\lambda$, and $2.0\text{m}$ for $L$ in expression (I)

  $\begin{array}{c}\left|dn\right|=\left|\left(\frac{2\left(2.0\text{m}\right)}{{\left(2\times {10}^{-2}\text{m}\right)}^2}\right)\right|\left|0.1\times {10}^{-2}\text{m}\right|\\ =10\end{array}$

The expression obtained the same value as found in sub part (a) that is the number of modes with wavelength between $2.0\text{cm}$ to $2.1\text{cm}$ is $10$.

(d)

To determine

The reason to justify replacing $\left|\left(\frac{\Delta n}{L\Delta \lambda }\right)\right|$ with $\left|\left(\frac{dn}{Ld\lambda }\right)\right|$

Answer to Problem 7P

For the shorter wavelengths $n$ is a continuous function of $\lambda$, $n=\frac{2L}{\lambda }$ is a discrete function.

Explanation of Solution

Write the expression the number of modes per unit wavelength per unit length,

  $\frac{\Delta n}{L\Delta \lambda }=\frac{\Delta n}{L\left(d\lambda \right)}$        (I)

Here, $n$ is the length of the string, and $L$ is the length.

Considering $\Delta \lambda$ tends to zero, $\frac{\Delta n}{\Delta \lambda }$ tends to $\frac{dn}{d\lambda }$ therefore,

  $\underset{\Delta \lambda \to 0}{\lim }\left|\frac{\Delta n}{L\Delta \lambda }\right|=\left|\frac{\Delta n}{L\left(d\lambda \right)}\right|$        (I)

Conclusion:

For the shorter wavelengths $\frac{\Delta n}{\Delta \lambda }$ can be replaced by $\frac{dn}{d\lambda }$.

And also for the shorter wavelengths $n$ is a continuous function of $\lambda$, $n=\frac{2L}{\lambda }$ is a discrete function.