91影视

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,$\mathrm{v}=344\text{鈥尘/蝉}$
3. Frequency of audio oscillator varies from$1000\text{Hzto}2000\text{Hz}.$

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, $\mathrm{f}=\mathrm{n}\frac{\mathrm{v}}{2\mathrm{L}}$ 鈥�(颈)

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

$\begin{array}{c}\mathrm{f}=\mathrm{n}\frac{344\text{鈥尘/蝉}}{2脳0.457\text{m}}\\ =\mathrm{n}\frac{344}{0.914}\frac{1}{\text{s}}\\ =\mathrm{n}(3.7636脳{10}^2)\text{Hz}\\ =376.36\mathrm{n}\text{Hz}\\ 鈮�376.4\mathrm{n}\text{Hz}\\ (\mathrm{n}\text{canbe}3,4,5\text{tomatchthefrequencyrangeof1000Hzto2000Hz})\end{array}$

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

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:

$\begin{array}{c}\mathrm{f}=3脳376.4\text{Hz}\\ =1129\text{鈥塇锄}\end{array}$

Hence, the value of lowest frequency is$1129\text{鈥塇锄}$

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:

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

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