Question

Pipe A, which is $\mathbf{1}\mathbf{.20}\mathbf{}\mathbf{\text{m}}$long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is$\mathbf{343}\mathbf{}\mathbf{\text{m/s}}$. Pipe$\mathbf{\text{B}}$, which is closed at one end, oscillates at its second lowest harmonic frequency. This frequency of$\mathbf{\text{B}}$happens to match the frequency of$\mathbf{\text{A}}$. An$\mathit{x}\mathbf{}\mathbf{\text{\hspace{0.17em}axis}}$extend along the interior of$\mathbf{\text{B}}$, with$\mathit{x}\mathbf{=}\mathbf{}\mathbf{0}$at the closed end. (a) How many nodes are along that axis? What is the (b) smallest and (c) second smallest value of$\mathit{x}$locating those nodes? (d) What is the fundamental frequency of$\mathit{B}$?

Short answer

1. Number of nodes along the x-axis is 2.
2. Smallest value of x locating those nodes is 0.
3. Second smallest value of x locating those nodes is $0.40\text{\hspace{0.17em}m}$.
4. Fundamental frequency of B is $143\text{\hspace{0.17em}Hz}$.

Step-by-step solution

Identification of given data

1. Speed of sound$\mathrm{v}=343\text{m/s}$
2. Length of pipe${\mathrm{L}}_{\mathrm{A}}=1.20\text{\hspace{0.17em}m}$

Significance of frequency

The number of waves passing a fixed location in a unit of time is referred to as frequency in physics.

It can be seen that “third lowest … frequency” corresponds to harmonic number${\mathbf{n}}_{\mathbf{A}}\mathbf{=}\mathbf{3}$for pipe A, which is open at both ends. Also, “second-lowest … frequency” corresponds to harmonic number${\mathbf{n}}_{\mathbf{B}}\mathbf{}\mathbf{=}\mathbf{3}$for pipe B, which is closed at one end. It means that the frequency of B matches the frequency of A.

Formula:

The frequency of body with n number of oscillations, $\mathrm{f}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}$ …(i)

(a) Determining the number of nodes 

As frequency of B matches the frequency of A, using equation (i) we can write that:

$\begin{array}{c}{\mathrm{f}}_{\mathrm{A}}={\mathrm{f}}_{\mathrm{B}}\\ 3\frac{\mathrm{v}}{2{\mathrm{L}}_{\mathrm{A}}}=3\frac{\mathrm{v}}{4{\mathrm{L}}_{\mathrm{B}}}\\ {\mathrm{L}}_{\mathrm{B}}=\frac{{\mathrm{L}}_{\mathrm{A}}}{2}\\ =0.60\text{\hspace{0.17em}m}\end{array}$

Now, the wavelength in terms of frequency is given as:

$\begin{array}{c}\mathrm{\lambda }=\frac{4{\mathrm{L}}_{\mathrm{B}}}{3}\\ =4\frac{0.60\text{\hspace{0.17em}m}}{3}\\ =0.8\text{\hspace{0.17em}m}\end{array}$

The change from node to anti-node requires a distance of $\mathrm{\lambda }/4$so that every increment of $0.20\text{m}$along the x axis involves a switch between node and anti-node. Since the closed end is a node, the next node appears at $\mathrm{x}=0.40\text{m}$, so there are 2 nodes.

(b) Determining the smallest value of x 

From part (a), the smallest value of x where the node is present is$\mathrm{x}=0$

(c) Determining the second smallest value of x

From part (a), the second smallest value of x where the node is present is$0.40\text{\hspace{0.17em}m}$

(d) Determining the fundamental frequency of B

Using the velocity of sound,$\mathrm{v}=343\text{m/s}$, we can find the frequency.

$\begin{array}{c}{\mathrm{f}}_3=\frac{\mathrm{v}}{\mathrm{\lambda }}\\ =\frac{343\text{\hspace{0.17em}m/s}}{0.8\text{\hspace{0.17em}m}}\\ =428.75\text{\hspace{0.17em}Hz}\end{array}$

Using this frequency, we can find the fundamental frequency as n=3,