
The air column in a pipe which is closed at one end will be in resonance with a vibrating tuning fork at a frequency 260Hz, if the length of the air column is (speed of sound in air=$330\dfrac{m}{{\sec }}$ )
A) 31.73 cm
B) 62.5 cm
C) 35.75 cm
D) 12.5 cm
Answer
587.7k+ views
Hint:Musical wind instruments like flute, clarinet etc. are based on the principle of vibrations of air columns. In a closed organ pipe, then the air column vibrates (as shown in figure) in the fundamental mode. There is a node at the closed end and an anti node at the open end. If l is the length of the tube then CoP(close organ pipe) formula is: $\dfrac{V}{{4l}}$ =f(first overtone),$\dfrac{{3V}}{{4l}}$=f (second overtone) and so on. V is the velocity and l is the length and f is frequency.
Step by step solution:
Step 1:
First we need to know about the close organ pipe (CoP): The organ pipe in which one end is opened and another end is closed is called organ pipe. Bottle, whistle, etc. are examples of closed organ pipes.
Therefore the frequency of ${p_{th}}$ overtone is (2p + 1) ${n_1}$ where${n_1}$ is the fundamental frequency. In a closed pipe only odd harmonics are produced. The frequencies of harmonics are in the ratio of 1 : 3 : 5 and so on.
Step 2:
The formula for the overtones in closed organ pipe is: $\dfrac{V}{{4l}}$=f (first overtone),$\dfrac{{3V}}{{4l}}$=f (second overtone) and so on. V is the velocity and l is length and f is frequency
So, we are given in question: a vibrating tuning fork at a frequency 260Hz
And speed of sound in air is $330\dfrac{m}{s}$
putting into the formula of overtone we will get $\dfrac{{n \times V}}{{4l}}$ =260, here n is the number of overtones.
Then for length substituting the values =$\dfrac{{n \times 330}}{{4 \times 260}}$ here, n=1
So the length for n=1 is 0.3173 m
Converting it into centimeter=0.3173×1000
This is equal to 31.73cm.
Option A. is correct
Note:In a closed pipe at one end, the wave takes 0.01 seconds to travel from closed end to open end. The frequency of the fundamental note of a tube closed at one end is 200Hz. A pipe is one end made to vibrate in its second overtone by a tuning fork of frequency 440Hz.
Unlike the other instrument types, there is no second harmonic for a closed end air column. The next frequency above the fundamental frequency is the third harmonic. In fact a closed end instrument does not possess an even numbered harmonics
Step by step solution:
Step 1:
First we need to know about the close organ pipe (CoP): The organ pipe in which one end is opened and another end is closed is called organ pipe. Bottle, whistle, etc. are examples of closed organ pipes.
Therefore the frequency of ${p_{th}}$ overtone is (2p + 1) ${n_1}$ where${n_1}$ is the fundamental frequency. In a closed pipe only odd harmonics are produced. The frequencies of harmonics are in the ratio of 1 : 3 : 5 and so on.
Step 2:
The formula for the overtones in closed organ pipe is: $\dfrac{V}{{4l}}$=f (first overtone),$\dfrac{{3V}}{{4l}}$=f (second overtone) and so on. V is the velocity and l is length and f is frequency
So, we are given in question: a vibrating tuning fork at a frequency 260Hz
And speed of sound in air is $330\dfrac{m}{s}$
putting into the formula of overtone we will get $\dfrac{{n \times V}}{{4l}}$ =260, here n is the number of overtones.
Then for length substituting the values =$\dfrac{{n \times 330}}{{4 \times 260}}$ here, n=1
So the length for n=1 is 0.3173 m
Converting it into centimeter=0.3173×1000
This is equal to 31.73cm.
Option A. is correct
Note:In a closed pipe at one end, the wave takes 0.01 seconds to travel from closed end to open end. The frequency of the fundamental note of a tube closed at one end is 200Hz. A pipe is one end made to vibrate in its second overtone by a tuning fork of frequency 440Hz.
Unlike the other instrument types, there is no second harmonic for a closed end air column. The next frequency above the fundamental frequency is the third harmonic. In fact a closed end instrument does not possess an even numbered harmonics
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