End Correction
End correction is a concept in physics that arises when studying the resonance of air columns in tubes, particularly in instruments like organ pipes and wind instruments. When sound waves travel through a tube and reflect at its open end, they don't abruptly stop at the physical end of the tube. Instead, the air just outside the tube vibrates slightly, extending the effective length of the tube. This extension is known as the end correction. In real life, this phenomenon is crucial for accurately tuning musical instruments. For instance, in a flute, the sound pitch would be slightly off if the end correction wasn't considered, leading to notes that are out of tune. Understanding and applying end correction helps ensure that the instrument produces the desired notes accurately, making it essential for musicians and engineers alike. In this article, we will cover the concept of End Correction this concept falls under the broader category of Oscillations and Waves.
End Correction
End correction is a phenomenon observed in the resonance of air columns within tubes, where the actual point of reflection of sound waves occurs slightly beyond the physical open end of the tube. This happens because the air particles at the open end of the tube continue to vibrate and extend the effective length of the air column.
In the organ pipe as we have studied in the last concept, when the wave reaches the open end, due to collision particles scatter away from the pipe. Due to this the density reduces outside the pipe and forms a rarer medium.
So, we can say that the wave is not exactly reflected back from the open end of the pipe. So, we can say that the antinodes will always form a little away from the open ends. We can see this in the given figure. So the distance above the open end where an antinode is formed is called end correction.
This end correction varies with the radius of the pipe and is given as $=e=0.6 r$
Now taking the end correction into account, the frequency of a closed pipe of length $l$ can be given as
$n_o=\frac{\nu}{4(l+e)}$ (One end open)
For open pipe
$n_o=\frac{\nu}{2(l+2 e)}$ (Both ends open)
Solved Examples Based on End Correction
Example 1: In a resonance pipe, the first and 2nd resonance are obtained at depts 22.7 cm & 70.2 cm. What will be the end correction (in cm)
1) 1.05
2) 1115.5
3) 92.5
4) 113.5
Solution:
Fundamental frequency with end correction
$\begin{aligned} & \nu_0=\frac{V}{4(l+e)} \quad \text { (one end open) } \\ & \nu_0=\frac{V}{2(l+2 e)} \quad \text { (Both ends open) } \\ & \mathrm{e}=\text { end correction }\end{aligned}$
For end correction
$\begin{aligned} & \frac{l_2+e}{l_1+e}=\frac{3 \lambda / 4}{\lambda / 4}=3 \\ & \mathrm{e}=\frac{l_2-3 l_1}{2}=\frac{70.2-3 \times 22.7}{2} \\ & \mathrm{e}=1.05 \mathrm{~cm}\end{aligned}$
Hence, the answer is the option (1).
Example 2: In resonance pipe in the fundamental mode, with a tuning fork is 0.1 m. When this length is changed to 0.35 m, the same tuning fork resonates with the first overtone. Calculate end correction
1) 0.012 m
2) 0.025 m
3) 0.05 m
4) 0.024 m
Solution:
Fundamental frequency with end correction
$$
\begin{aligned}
& \nu_0=\frac{V}{4(l+e)} \quad \text { (one ends open) } \\
& \nu_0=\frac{V}{2(l+2 e)} \quad \text { (Both ends open) } \\
& e=\text { end correction }
\end{aligned}
$$
Let e be the end correction
$$
\begin{aligned}
& \frac{v}{4\left(l_1+e\right)}=\frac{3 v}{4\left(l_2+e\right)} \Rightarrow \\
& e=2.5 \mathrm{~cm}=0.025 \mathrm{~m}
\end{aligned}
$$
Hence, the answer is the option (2).
Example 3: In a resonance column, two successive resonances are obtained at depths 30 cm and 50 cm respectively. The next resonance will be obtained at a depth
1) 60 cm
2) 80 cm
3) 70 cm
4) 90 cm
Solution:
Given, $\frac{\lambda}{2}=(50-30) \mathrm{cm}=20 \mathrm{~cm}$
Neat resonance will be obtained at $50+\frac{\lambda}{2}$
$$
=70 \mathrm{~cm} \text {. }
$$
Hence, the answer is the option (3).
Example 4: A pipe closed at one end produces a fundamental note of 412 Hz. It's cut into two pieces of equal length the fundamental notes produced by the two pieces are
1) $824 \mathrm{~Hz}, 1648 \mathrm{~Hz}$
2) $412 \mathrm{~Hz}, 824 \mathrm{~Hz}$
3) $206 \mathrm{~Hz}, 412 \mathrm{~Hz}$
4) $206 \mathrm{~Hz}, 824 \mathrm{~Hz}$.
Solution:
$\begin{aligned} & \mathrm{f}_1=\frac{\mathrm{v}}{2(\mathrm{l} / 2)}=\frac{\mathrm{v}}{\mathrm{l}}=4 \times 412=1648 \mathrm{~Hz} \\ & \mathrm{f}_2=\frac{\mathrm{v}}{4(\mathrm{l} / 2)}=\frac{\mathrm{v}}{2 \mathrm{l}}=2 \times 412=824 \mathrm{~Hz}\end{aligned}$
Hence, the answer is the option (1).
Summary