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Question: An air column in a pipe, which is closed at one end, is in resonance with a vibrating tuning fork of...

An air column in a pipe, which is closed at one end, is in resonance with a vibrating tuning fork of frequency 364 Hz. If v=330m/sv=330m/s , the length of the column in cm is (are)\left( \text{are} \right) :
A. 3125 B. 62.50 C. 93.75 D. 125 \begin{aligned} & \text{A}\text{. 3125} \\\ & \text{B}\text{. 62}\text{.50} \\\ & \text{C}\text{. 93}\text{.75} \\\ & \text{D}\text{. 125} \\\ \end{aligned}

Explanation

Solution

When a vibrating tuning fork is held horizontally near the open end of a closed pipe, sound waves are sent by source (like a vibrating tuning fork) near the open end. When waves pave in pipe, nodes and antinodes are produced in pipe. Number of nodes & antinodes depends on the length of pipe & hence nodes are formed. Use the formula or expression which gives relations between mode, harmonics, length & wavelength.

Formula used:
Length of air column is given by –
l=(2n+1)d4l=\left( 2n+1 \right)\dfrac{d}{4}
Where, n=0,1,2,3,.....n=0,1,2,3,.....
d = wavelength
l = length of pipe.

Complete step by step answer:
We have given an air column in a pipe which is closed at one end. A frequency of tuning fork is given as:
f=264Hzf=264Hz
Velocity of wave in pipe is given as –
v=330m/sv=330m/s
To calculate wavelength, we have –
v=fd d=vf =330264 d=1.25m \begin{aligned} & v=fd \\\ & \Rightarrow d=\dfrac{v}{f} \\\ & =\dfrac{330}{264} \\\ & d=1.25m \\\ \end{aligned}
Wavelength is equal to 1.25m
We know that there are certain discrete frequencies with which standing waves can be setup in a closed pipe, such frequency is known as natural frequency.
In the fundamental mode of vibration of the air column closed at one end, only one node and only antinode present.
Therefore, length of air column is given as –
l=(2n+r)d4l=\left( 2n+r \right)\dfrac{d}{4}
We know that, for first mode or fundamental mode, value of n=0n=0
l=(2×0+1)1.254=1.254m=0.3125m=31.25cm\therefore l=\left( 2\times 0+1 \right)\dfrac{1.25}{4}=\dfrac{1.25}{4}m=0.3125m=31.25cm
Therefore, the length of the column is 31.25cm. Option (A) is the correct option.
In the second mode of the vibration of the air column closed at one end, only two nodes & two antinodes are present.
Therefore, length of air column is given as –
l=(2n+1)d4l=\left( 2n+1 \right)\dfrac{d}{4}
We know that, for second node, n=1n=1
l=(2×1+1)1.254=3×1.254m=93.75cm.\therefore l=\left( 2\times 1+1 \right)\dfrac{1.25}{4}=\dfrac{3\times 1.25}{4}m=93.75cm.
In third mode, n=2n=2
l=(2×2+1)1.254=5×1.254m=156.25cm\therefore l=\left( 2\times 2+1 \right)\dfrac{1.25}{4}=\dfrac{5\times 1.25}{4}m=156.25cm
**Therefore, (A) & (C) are correct options. **

Note:
Wavelength travel in pipe closed one end is quarter of wavelength while wavelength travel in pipe closed at both ends is half of the wavelength. Frequency of fundamental mode is the coolest frequency. Second mode is also known as third harmonic or first overtone, whereas third mode is known as fifth harmonic or second overtone. Note that only odd harmonics are present as overtone, in a mode of vibration of air column which is closed at one end.