The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe iscm.

Question

The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20 cm. The length of the open organ pipe iscm. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f 1=(2 v /2 l 1) f 2=( v /4 l 2) f 1= f 2 =(2 v /2 l 1)=( v /4 l 2) l 1=4 l 2=80 cm

Q. The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is $20 c m$ . The length of the open organ pipe is_______ $c m$ .

$f_{1} = \frac{2 v}{2 l _{1}}$
$f_{2} = \frac{v}{4 l _{2}}$
$f_{1} = f_{2}$
$= \frac{2 v}{2 l _{1}} = \frac{v}{4 l _{2}}$
$l_{1} = 4 l_{2} = 80 c m$

Q. The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is 20cm. The length of the open organ pipe is_______cm.

f1​=2l1​2v​ f2​=4l2​v​ f1​=f2​ =2l1​2v​=4l2​v​ l1​=4l2​=80cm

Q. The first overtone frequency of an open organ pipe is equal to the fundamental frequency of a closed organ pipe. If the length of the closed organ pipe is $20 c m$ . The length of the open organ pipe is_______ $c m$ .

$f_{1} = \frac{2 v}{2 l _{1}}$
$f_{2} = \frac{v}{4 l _{2}}$
$f_{1} = f_{2}$
$= \frac{2 v}{2 l _{1}} = \frac{v}{4 l _{2}}$
$l_{1} = 4 l_{2} = 80 c m$