Question

A tube of length L is open at both ends. A stationary wave is set up in this tube when a tuning fork vibrating with frequency fx is held at one end. This is the lowest frequency of stationary wave that can be formed in this tube. Another tube of length 2L is closed at one end. A stationary wave is set up in this tube when a tuning fork vibrating with frequency fy is held at the open end. This is the lowest frequency of stationary wave that can be formed in this tube. Assume the end correction for each tube is negligible.
Which equation is correct?
A fx =4fy
B fx =2fy
C fx = 2fy
D fx = 4fy

Answer 1

Answer: like it and thanks

Explanation:

Answer 2

Given :

Length of first tube = L,

Length of second tube = 2L.

Frequency in first tube = fx.

Frequency in second tube = fy.

To find:

Relation between fx and fy.

Solution :

We know that,

Velocity of stationary wave in both cases is same.

Let velocity of stationary wave :

$= v \: \: \frac{m}{s}$

In case of first tube,

Length of tube = L,

and it is given that in this case the tube is

Open from both ends.

So frequency of wave in case of open end tube is equal to :

$f = \frac{velocity}{(2 \times length \: of \: tube)}$

So in this case:

Frequency of first tube fx :

$fx = \frac{v}{2L} \:$

In case of second tube,

Length of tube = 2L,

and it is given that in this case the tube is

Closed from one end.

So frequency of wave in case of close end tube is equal to :

$f = \frac{velocity}{(4\times length \: of \: tube )\: } \:$

So in this case:

Frequency of second tube fy :

$fy = \frac{v}{(4 \times (2L ))\: } = \frac{fx}{4}$

so,

fx = 4 fy

So the correct option is A and D.