In a pipe open at both ends, ‘n1’ and ‘n2’ be the frequencies corresponding to vibrating lengths ‘l1’ and ‘l2’ respectively. The end correction is

Question

In a pipe open at both ends, ‘n1’ and ‘n2’ be the frequencies corresponding to vibrating lengths ‘l1’ and ‘l2’ respectively. The end correction is (A) (n1l1-n2l2/2(n1-n2)) (B) (n2l2-n1l1/2(n2-n1)) (C) (n2l2-n1l1/2(n1-n2)) (D) (n1l1-n2l2/(n1-n2))

Tardigrade Admin · Accepted Answer

The frequency of the pth mode of vibration in an open organ pipe, v = (pv/2(L + 2e)) Where, e is the end correction of one side and L is the length of the pipe Given: For L = l1, frequency of pth mode = n1 For L = l2, frequency of pth mode = n2 ⇒ n1 = (pv/2(l1 + 2e)) dots(1) ⇒ n2 = (pv/2(l2 + 2e)) dots (2) Dividing (1) and (2), we get (n1/n2) = (pv/2(l1 + 2e)) × (2(l2 + 2e)/pv) ∴ (n1/n2) = (l2 + 2e/l1 + 2e) ⇒ e = (n2l2 - n1l1/2(n1 - n2))

Q. In a pipe open at both ends, ‘ $n_{1}$ ’ and ‘ $n_{2}$ ’ be the frequencies corresponding to vibrating lengths ‘ $l_{1}$ ’ and ‘ $l_{2}$ ’ respectively. The end correction is

$\frac{n _{1} l _{1} - n _{2} l _{2}}{2 ( n _{1} - n _{2} )}$

$\frac{n _{2} l _{2} - n _{1} l _{1}}{2 ( n _{2} - n _{1} )}$

$\frac{n _{2} l _{2} - n _{1} l _{1}}{2 ( n _{1} - n _{2} )}$

$\frac{n _{1} l _{1} - n _{2} l _{2}}{( n _{1} - n _{2} )}$

Q. In a pipe open at both ends, ‘n1​’ and ‘n2​’ be the frequencies corresponding to vibrating lengths ‘l1​’ and ‘l2​’ respectively. The end correction is

2(n1​−n2​)n1​l1​−n2​l2​​

2(n2​−n1​)n2​l2​−n1​l1​​

2(n1​−n2​)n2​l2​−n1​l1​​

(n1​−n2​)n1​l1​−n2​l2​​

Q. In a pipe open at both ends, ‘ $n_{1}$ ’ and ‘ $n_{2}$ ’ be the frequencies corresponding to vibrating lengths ‘ $l_{1}$ ’ and ‘ $l_{2}$ ’ respectively. The end correction is

$\frac{n _{1} l _{1} - n _{2} l _{2}}{2 ( n _{1} - n _{2} )}$

$\frac{n _{2} l _{2} - n _{1} l _{1}}{2 ( n _{2} - n _{1} )}$

$\frac{n _{2} l _{2} - n _{1} l _{1}}{2 ( n _{1} - n _{2} )}$

$\frac{n _{1} l _{1} - n _{2} l _{2}}{( n _{1} - n _{2} )}$