
A standing wave propagating with velocity $ 300m/s $ in an open pipe of length $ 4m $ has four nodes. The frequency of the wave is?
(A) $ 75Hz $
(B) $ 100Hz $
(C) $ 150Hz $
(D) $ 300Hz $
(E) $ 600Hz $
Answer
467.7k+ views
Hint : To solve this question, we need to find out the length of the pipe in terms of the wavelength with the help of the number of nodes given in the question. For this we have to draw the diagram of the nodes and antinodes in an open pipe and from that we can easily find out the relation. Then, we have to use the relation between the frequency and the wavelength to get the final answer.
Formula Used: The formula which is used to solve this question is given by
$\Rightarrow \lambda f = v $ , here $ \lambda $ is the wavelength, $ f $ is the frequency, and $ v $ is the velocity of a wave.
Complete step by step answer
Let the frequency of the wave be $ f $ .
According to the question, we have an open pipe in which a standing wave has four nodes. So, this situation can be represented by the below diagram.
As we know that the distance between a node and an antinode is equal to one fourth of the wavelength of the light, so the total length of the pipe becomes
$\Rightarrow L = 8\left( {\dfrac{\lambda }{4}} \right) $
$ \Rightarrow L = 2\lambda $
Dividing both the sides by $ 2 $ , we get
$\Rightarrow \lambda = \dfrac{L}{2} $
According to the question, the length of the pipe is $ 4m $ . Therefore substituting $ L = 4m $ above we get
$\Rightarrow \lambda = \dfrac{4}{2} = 2m $ (1)
So the wavelength of the wave is equal to $ 2m $ .
Now, we know that the frequency and the wavelength of a wave are related by the relation
$\Rightarrow \lambda f = v $
So the frequency of the wave is given by
$\Rightarrow f = \dfrac{v}{\lambda } $
According to the question, $ v = 300m/s $ . Also, from (1) we have $ \lambda = 2m $ . Substituting these above we get
$\Rightarrow f = \dfrac{{300}}{2} $
$ \Rightarrow f = 150Hz $
Thus the frequency of the wave is equal to $ 150Hz $ .
Hence, the correct answer is option C.
Note
We need to be clear with the terminology of the organ pipes. An open organ pipe is a pipe which is closed at both the ends. While a closed organ pipe is closed only at one end. It must be noted that at least one end of the organ pipe must be open.
Formula Used: The formula which is used to solve this question is given by
$\Rightarrow \lambda f = v $ , here $ \lambda $ is the wavelength, $ f $ is the frequency, and $ v $ is the velocity of a wave.
Complete step by step answer
Let the frequency of the wave be $ f $ .
According to the question, we have an open pipe in which a standing wave has four nodes. So, this situation can be represented by the below diagram.

As we know that the distance between a node and an antinode is equal to one fourth of the wavelength of the light, so the total length of the pipe becomes
$\Rightarrow L = 8\left( {\dfrac{\lambda }{4}} \right) $
$ \Rightarrow L = 2\lambda $
Dividing both the sides by $ 2 $ , we get
$\Rightarrow \lambda = \dfrac{L}{2} $
According to the question, the length of the pipe is $ 4m $ . Therefore substituting $ L = 4m $ above we get
$\Rightarrow \lambda = \dfrac{4}{2} = 2m $ (1)
So the wavelength of the wave is equal to $ 2m $ .
Now, we know that the frequency and the wavelength of a wave are related by the relation
$\Rightarrow \lambda f = v $
So the frequency of the wave is given by
$\Rightarrow f = \dfrac{v}{\lambda } $
According to the question, $ v = 300m/s $ . Also, from (1) we have $ \lambda = 2m $ . Substituting these above we get
$\Rightarrow f = \dfrac{{300}}{2} $
$ \Rightarrow f = 150Hz $
Thus the frequency of the wave is equal to $ 150Hz $ .
Hence, the correct answer is option C.
Note
We need to be clear with the terminology of the organ pipes. An open organ pipe is a pipe which is closed at both the ends. While a closed organ pipe is closed only at one end. It must be noted that at least one end of the organ pipe must be open.
Recently Updated Pages
The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

A certain household has consumed 250 units of energy class 11 physics CBSE

The lightest metal known is A beryllium B lithium C class 11 chemistry CBSE

What is the formula mass of the iodine molecule class 11 chemistry CBSE

Trending doubts
Why was the Vernacular Press Act passed by British class 11 social science CBSE

Arrange Water ethanol and phenol in increasing order class 11 chemistry CBSE

Name the nuclear plant located in Uttar Pradesh class 11 social science CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

How did silk routes link the world Explain with three class 11 social science CBSE

What are the various challenges faced by political class 11 social science CBSE
