Answer
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Hint: They’ve given an air column closed at one end which makes it an open organ pipe. In order to find the length of the air column, we must first find the wavelength of the sound wave formed from the tuning fork. Using this frequency we can determine the condition for each resonance.
Formula used:
$\lambda = \dfrac{v}{f}$
${l_1} = \dfrac{\lambda }{4}$
${l_2} = \dfrac{{3\lambda }}{4}$
Complete answer:
The wavelength of the sound wave is given by
$\lambda = \dfrac{v}{f}$
Here,
$\lambda $ is the wavelength of the sound wave
$v$ is the speed of the sound
$f$ is the frequency of the sound wave
The velocity and the frequency of the sound wave are given in the question. Substituting these values, we are going to get
$\eqalign{
& \lambda = \dfrac{v}{f} \cr
& \Rightarrow \lambda = \dfrac{{330}}{{264}} \cr
& \Rightarrow \lambda = 1.25m \cr} $
In the question, they’ve given an air column pipe which is closed at one end whose air column length should be found, so that it is in resonance with a tuning fork of frequency 264Hz. The pipe is a closed organ pipe. So, it will be in resonance with the tuning fork when the length of the column, $l$ is such that
$l = \dfrac{\lambda }{4},\dfrac{{3\lambda }}{4},\dfrac{{5\lambda }}{4}..........$
Applying the condition for first resonance the length of the air column will be
${l_1} = \dfrac{\lambda }{4} = \dfrac{{1.25}}{4} = 0.3125m = 31.25cm$
Similarly applying the condition for second resonance we will have
${l_2} = \dfrac{{3\lambda }}{4} = \dfrac{{3 \times 1.25}}{4} = 0.9375m = 93.75cm$
So, the correct answer is “Option A and C”.
Note:
The length of the pipe will be in multiples of $\dfrac{\lambda }{4}$ as it is the distance between a node and an antinode. A closed organ pipe is that which is closed at one end. In this case, a node is formed at the closed end. On the other hand, an open organ pipe has open ends at which antinodes are formed.
Formula used:
$\lambda = \dfrac{v}{f}$
${l_1} = \dfrac{\lambda }{4}$
${l_2} = \dfrac{{3\lambda }}{4}$
Complete answer:
The wavelength of the sound wave is given by
$\lambda = \dfrac{v}{f}$
Here,
$\lambda $ is the wavelength of the sound wave
$v$ is the speed of the sound
$f$ is the frequency of the sound wave
The velocity and the frequency of the sound wave are given in the question. Substituting these values, we are going to get
$\eqalign{
& \lambda = \dfrac{v}{f} \cr
& \Rightarrow \lambda = \dfrac{{330}}{{264}} \cr
& \Rightarrow \lambda = 1.25m \cr} $
In the question, they’ve given an air column pipe which is closed at one end whose air column length should be found, so that it is in resonance with a tuning fork of frequency 264Hz. The pipe is a closed organ pipe. So, it will be in resonance with the tuning fork when the length of the column, $l$ is such that
$l = \dfrac{\lambda }{4},\dfrac{{3\lambda }}{4},\dfrac{{5\lambda }}{4}..........$
Applying the condition for first resonance the length of the air column will be
${l_1} = \dfrac{\lambda }{4} = \dfrac{{1.25}}{4} = 0.3125m = 31.25cm$
Similarly applying the condition for second resonance we will have
${l_2} = \dfrac{{3\lambda }}{4} = \dfrac{{3 \times 1.25}}{4} = 0.9375m = 93.75cm$
So, the correct answer is “Option A and C”.
Note:
The length of the pipe will be in multiples of $\dfrac{\lambda }{4}$ as it is the distance between a node and an antinode. A closed organ pipe is that which is closed at one end. In this case, a node is formed at the closed end. On the other hand, an open organ pipe has open ends at which antinodes are formed.
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