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Sagot :
Given:
Length of pipe = 1.44 m
Frequency of second pipe = 1.3 Hz
Let's find the difference between the lengths of the pipes.
Here, we have:
[tex]f_{beat}=f_1-f_2[/tex]Thus, we have:
[tex]f_2=f-1.3[/tex]To find the frequency of the pipe, f,, we have:
[tex]\begin{gathered} f=\frac{v}{4l} \\ \\ Where: \\ v\text{ is the speed of sound = 343 m/s} \\ \\ f=\frac{343}{4*1.44} \\ \\ f=59.55\text{ Hz} \end{gathered}[/tex]Plug in the value of f and find f2:
[tex]\begin{gathered} f_2=59.55-1.3 \\ \\ f_2=58.25\text{ Hz} \end{gathered}[/tex]Now, let's find the length of the second pipe:
[tex]\begin{gathered} f=\frac{v}{4l} \\ \\ l_2=\frac{v}{4f_2} \\ \\ l_2=\frac{343}{4*58.25} \\ \\ l_2=1.47\text{ m} \end{gathered}[/tex]Therefore, the difference in length will be:
L2 - L1 = 1.47 m - 1.44 m = 0.03 m
Therefore, the second pipe will be longer by 0.03 meters.
ANSWER:
0.03 m
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