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
Verified
459k+ 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
Master Class 11 Accountancy: Engaging Questions & Answers for Success
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Trending doubts
10 examples of friction in our daily life
Difference Between Prokaryotic Cells and Eukaryotic Cells
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE
Define least count of vernier callipers How do you class 11 physics CBSE