Question

A loudspeaker that produces signals from $50$ to $500\mathrm{Hz}$ is placed at the open end of a closed tube of length $1.1\mathrm{m}$. What are the lowest and the highest frequencies that excite resonance?

Answer 1

Step 1: Given data

Length of the tube $=1.1\mathrm{m}$

Velocity of sound = $330\mathrm{m}{\sec }^{-1}$

Step 2: Formula used

The resonant frequencies for the tube closed at one end are then given by:

${\mathrm{f}}_{\mathrm{n}}=\frac{\mathrm{nv}}{4\mathrm{L}}$ ………………….(1)

Where, L is the length of the tube

$\mathrm{\nu }$ is the velocity of sound

$\mathrm{n}=1,3,5,...$

Step 3: Find the value of $\mathrm{n}$

Value of $\mathrm{n}$, for higher frequencies:

Substituting the value of $\mathrm{L},\mathrm{v}$ and ${\mathrm{f}}_{\mathrm{n}}$ in equation (1), we get

$$\begin{gathered} {\mathrm{f}}_{\mathrm{n}}=\frac{\mathrm{n}\times 330}{4\times 1.1}=500\mathrm{Hz} \\ \Rightarrow \mathrm{n}=6.666 \end{gathered}$$

Value of $\mathrm{n}$ for lower frequencies is $1$, as the minimum value of $\mathrm{n}$ is $1$.

Step 4: Find the lowest and the highest frequencies that excite resonance

Putting the value of $\mathrm{n}$ in equation (1), we get

The highest frequency,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{h}}=\frac{6\times 330}{4\times 1.1} \\ {\mathrm{f}}_{\mathrm{h}}=\frac{1980}{4.4}=450\mathrm{Hz} \end{gathered}$$
The lowest frequency,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{l}}=\frac{1\times 330}{1.1\times 4} \\ {\mathrm{f}}_{\mathrm{l}}=\frac{330}{4.4}=75\mathrm{Hz} \end{gathered}$$

Thus, the highest and the lowest frequencies that excite resonance are $450\mathrm{Hz}$ and $75\mathrm{Hz}$.