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GalinKa [24]
3 years ago
5

A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at on

e end or not. Assume the speed of sound to be 340 m/s.
Engineering
1 answer:
MAXImum [283]3 years ago
5 0

Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

f_n=\frac{nv}{2L}

For (n+1)th value the frequency is

f_{n+1}=\frac{(n+1)v}{2L}

Taking a ratio of both equation and solving for n such that the value of n is a whole number

\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\

So n is a whole number this means that the pipe is open ended.

For confirmation the  nth frequency for a closed ended pipe is given as

f_n=\frac{(2n+1)v}{4L}

For (n+1)th value the frequency is

f_{n+1}=\frac{(2n+3)v}{4L}

Taking a ratio of both equation and solving for n such that the value of n is a whole number

\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m

The value of length is 3.4m.

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