91影视

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

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 鈥渢hird lowest 鈥� frequency鈥� corresponds to harmonic number${\mathbf{n}}_{\mathbf{A}}\mathbf{=}\mathbf{3}$for pipe A, which is open at both ends. Also, 鈥渟econd-lowest 鈥� frequency鈥� 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, $\mathrm{f}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}$ 鈥�(颈)

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

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

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

$\begin{array}{c}\mathrm{位}=\frac{4{\mathrm{L}}_{\mathrm{B}}}{3}\\ =4\frac{0.60\text{鈥尘}}{3}\\ =0.8\text{鈥尘}\end{array}$

The change from node to anti-node requires a distance of $\mathrm{位}/4$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.

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

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

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

$\begin{array}{c}{\mathrm{f}}_3=\frac{\mathrm{v}}{\mathrm{位}}\\ =\frac{343\text{鈥尘/s}}{0.8\text{鈥尘}}\\ =428.75\text{鈥塇锄}\end{array}$

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