The water level in a vertical glass tube $1m$ long can be adjusted to any position in the tube. A tuning fork vibrating at $686Hz$ is held just over the open top end of the tube to set up a standing wave of sound in the air filled top portion of the tube. (That air filled top portion acts as a tube with one end closed and the other end open)
(a) for how many different positions of the water lined will sound from the fork set up resonance in the tube’s air filled portion? What are the (b) least and
(c) Second least water height in the tube for resonance to occur?
Answer
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Hint: The tube partly filled water may be considered as one end closed. The glass tube shall behave as an organ pipe. I will behave as a closed end organ pipe and the length of the pipe is varied by varying the water level.
Formula used: $V=f\lambda $ Where $V$is velocity, $f$ if frequency.
$\lambda $ wavelength.
Complete step by step answer:
We are given the following data;
Frequency $f=686Hz$
Assume speed of sound to be ${{V}_{s}}=343\text{ m/s}$
Let $L$ be the length of the air column.
So, we have $h=1-L$
We know that, the resonance condition in tube is given by;
$\Rightarrow 4L=n\lambda $
Simplifying the above equation we get;
$\Rightarrow L=\dfrac{n\lambda }{4}$
And by the formula we have wavelength
$\Rightarrow \lambda =\dfrac{V}{f}=\dfrac{343}{686}=\dfrac{1}{2}=0.5\,\text{m}\text{.}$
for $n=1,{{L}_{1}}=\dfrac{\lambda }{4}=\dfrac{0.5}{4}=0.125\text{ m}$
for $n=2,{{L}_{2}}=\dfrac{3\lambda }{4}=\dfrac{3\times 0.5}{4}=0.375\text{ m}$
for $n=3,{{L}_{3}}=\dfrac{5\lambda }{4}=\dfrac{5\times 0.5}{4}=0.625\text{ m}$
$\because $ one end is fixed $n=1,3,5,7,9....$
(a) There are $4$ values of $n$ i.e. $n=1,3,5,7$ which satisfies $h>0$
(b) Smallest height of water for resonance to correspond to $n=7$ with $h=0.125\text{ m}\text{.}$
for $n=7,{{L}_{7}}=7\times 0.125=0.875$
Thus we have;
$h=1-{{L}_{7}}=1-875-0.125\text{ m}$
(c) The second smallest water height $h$ is at $n=5$
for $n=5,{{L}_{5}}=5\times 0.125=0.625\text{ m}$
thus we have;
$h=1-{{L}_{5}}=1-0.625=0.375\text{ m}$
Note: Two sound waves in opposite directions interfere with each other to create resonance. Waves produced are longitudinal stationary. Resonance tube is a partially filled hollow cylinder with water and forced into vibration by a running fork. The running fork is the object that forced the air inside of the resonance tube into resonance.
Formula used: $V=f\lambda $ Where $V$is velocity, $f$ if frequency.
$\lambda $ wavelength.
Complete step by step answer:
We are given the following data;
Frequency $f=686Hz$
Assume speed of sound to be ${{V}_{s}}=343\text{ m/s}$
Let $L$ be the length of the air column.
So, we have $h=1-L$
We know that, the resonance condition in tube is given by;
$\Rightarrow 4L=n\lambda $
Simplifying the above equation we get;
$\Rightarrow L=\dfrac{n\lambda }{4}$
And by the formula we have wavelength
$\Rightarrow \lambda =\dfrac{V}{f}=\dfrac{343}{686}=\dfrac{1}{2}=0.5\,\text{m}\text{.}$
for $n=1,{{L}_{1}}=\dfrac{\lambda }{4}=\dfrac{0.5}{4}=0.125\text{ m}$
for $n=2,{{L}_{2}}=\dfrac{3\lambda }{4}=\dfrac{3\times 0.5}{4}=0.375\text{ m}$
for $n=3,{{L}_{3}}=\dfrac{5\lambda }{4}=\dfrac{5\times 0.5}{4}=0.625\text{ m}$
$\because $ one end is fixed $n=1,3,5,7,9....$
(a) There are $4$ values of $n$ i.e. $n=1,3,5,7$ which satisfies $h>0$
(b) Smallest height of water for resonance to correspond to $n=7$ with $h=0.125\text{ m}\text{.}$
for $n=7,{{L}_{7}}=7\times 0.125=0.875$
Thus we have;
$h=1-{{L}_{7}}=1-875-0.125\text{ m}$
(c) The second smallest water height $h$ is at $n=5$
for $n=5,{{L}_{5}}=5\times 0.125=0.625\text{ m}$
thus we have;
$h=1-{{L}_{5}}=1-0.625=0.375\text{ m}$
Note: Two sound waves in opposite directions interfere with each other to create resonance. Waves produced are longitudinal stationary. Resonance tube is a partially filled hollow cylinder with water and forced into vibration by a running fork. The running fork is the object that forced the air inside of the resonance tube into resonance.
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