A tuning fork of frequency n is held near the open end of a tube, which is closed at the other end, and the length of the tube is adjusted until resonance occurs. If the two shortest lengths to produce resonance are L1 and L2 , then the speed of the sound is

Question

A tuning fork of frequency n is held near the open end of a tube, which is closed at the other end, and the length of the tube is adjusted until resonance occurs. If the two shortest lengths to produce resonance are L1 and L2 , then the speed of the sound is (A) n(.L2-L1.) (B) (n (L2 - L1)/2) (C) 4n(.L2-L1.) (D) 2n(.L2-L1.)

Tardigrade Admin · Accepted Answer

n= (v/4 L1) and n=(3 v/4 L2) ∴ v=4nL1 and 3v=4nL2 or 2v=4n(L2 - L1) or v=2n(L2 - L1)

Q. A tuning fork of frequency $n$ is held near the open end of a tube, which is closed at the other end, and the length of the tube is adjusted until resonance occurs. If the two shortest lengths to produce resonance are $L_{1}$ and $L_{2}$ , then the speed of the sound is

$n (L_{2} - L_{1})$

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

$4 n (L_{2} - L_{1})$

$2 n (L_{2} - L_{1})$

$n = \frac{v}{4 L _{1}}$ and $n = \frac{3 v}{4 L _{2}}$
$∴ v = 4 n L_{1}$ and $3 v = 4 n L_{2}$
$or 2 v = 4 n (L_{2} - L_{1})$
$or v = 2 n (L_{2} - L_{1})$

Q. A tuning fork of frequency n is held near the open end of a tube, which is closed at the other end, and the length of the tube is adjusted until resonance occurs. If the two shortest lengths to produce resonance are L1​ and L2​ , then the speed of the sound is

n(L2​−L1​)

2n(L2​−L1​)​

4n(L2​−L1​)

2n(L2​−L1​)