Question

In Figure, $\mathbf{S}$is a small loudspeaker driven by an audio oscillator with a frequency that is varied from$\mathbf{1000}\mathbf{}\mathbf{\text{\hspace{0.17em}Hz\hspace{0.17em}}}\mathbf{}\mathbf{\text{to}}\mathbf{}\mathbf{2000}\mathbf{\text{\hspace{0.17em}Hz}}$, and$\mathbf{D}$is a cylindrical pipe with two open ends and a length of$\mathbf{45.7}\mathbf{}\mathbf{\text{cm}}$. The speed of sound in the air-filled pipe is$\mathbf{344}\mathbf{}\mathbf{\text{m/s}}\mathbf{.}$ (a) At how many frequencies does the sound from the loudspeaker set up resonance in the pipe? What are the (b) lowest and (c) second lowest frequencies at which resonance occurs?

Short answer

1. The sound from the loudspeaker sets up resonance in the pipe at$376.4\text{\hspace{0.17em}nHz}$, where n=3,4 and 5
2. The lowest frequency at which resonance occurs is$1129\text{\hspace{0.17em}Hz}$
3. The second lowest frequency at which resonance occurs is$1506\text{\hspace{0.17em}Hz}$

Step-by-step solution

Identification of given data

1. Length of the air filled pipe,$\left(\mathrm{L}\right)=45.7\text{cmor}0.457\text{m}$
2. The speed of sound in air filled pipe,$\mathrm{v}=344\text{\hspace{0.17em}m/s}$
3. Frequency of audio oscillator varies from$1000\text{Hzto}2000\text{Hz}.$

Significance of frequency

The number of waves passing a fixed location in a unit of time is referred to as frequency in physics. It is given that the audio frequency of the loudspeaker is varied from 1000Hz to 2000Hz. We are given the length of pipe and speed of sound in the air-filled pipe. That is, we have to find out frequencies that lie between that given ranges.

Formula:

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

(a) Determining the frequency at which the loudspeaker sets up resonance

To find the frequency at which sound from the loudspeaker sets up resonance in the pipe, considering the formula from equation (i), we get

$\begin{array}{c}\mathrm{f}=\mathrm{n}\frac{344\text{\hspace{0.17em}m/s}}{2\times 0.457\text{m}}\\ =\mathrm{n}\frac{344}{0.914}\frac{1}{\text{s}}\\ =\mathrm{n}(3.7636\times {10}^2)\text{Hz}\\ =376.36\mathrm{n}\text{Hz}\\ \approx 376.4\mathrm{n}\text{Hz}\\ (\mathrm{n}\text{canbe}3,4,5\text{tomatchthefrequencyrangeof1000Hzto2000Hz})\end{array}$

Where n is varied as integral number. So, at that amount of frequency, the sound from the loudspeaker sets up resonance in the pipe.

Hence, the frequency value is given as$376.4\mathrm{n}\text{Hz}$ , where n=3,4 and 5

(b) Determining the lowest frequency at which resonance occurs  

In the above part (a), the frequency value can be seen as:

$376.4\mathrm{n}\text{Hz}$, where n=3,4 and 5

Hence, the lowest frequency of this range can be given for n = 3 as:

$\begin{array}{c}\mathrm{f}=3\times 376.4\text{Hz}\\ =1129\text{\hspace{0.17em}Hz}\end{array}$

Hence, the value of lowest frequency is$1129\text{\hspace{0.17em}Hz}$

(c) Determining the second lowest frequency

In the above part (a), the frequency value can be seen as:

$376.4\mathrm{n}\text{Hz}$, where n=3,4 and 5

Hence, the second lowest frequency of this range can be given for n = 4 as:

$\begin{array}{c}\mathrm{f}=4\times 376.4\text{Hz}\\ =1506\text{Hz}\end{array}$

So, the second lowest frequency is $1506\text{Hz}$