Question

Prove that for pipe closed at one end, the end correction is `e = (n_2l_2-n_1l_1)/(n_1-n_2)`

Answer 1

Consider the two pipes of same diameter but different lengths of l₁ and l₂·

Let n₁ and n₂ be the frequencies of tunning fork.

We know, for a pipe closed at one end

V = 4n₁L₁ = 4n₂L₂

n₁L₁ = n₂L₂

n₁(l₁ + e) = n₂(l₂ + e)

where, e = end correction

n₁l₁ + n₁e = n₂l₂ + n₂e

n₁e - n₂e = n₂l₂ - n₁l₁

e(n₁ - n₂) = n₂l₂ - n₁l₁

`therefore e = (n_2l_2-n_1l_1)/(n_1-n_2)`