Question

A student has a variable length air tube that is open at both ends and a $345Hz$ tuning fork. The student finds that the tube resonates with the tuning fork when the tube is a specific length. As the student slowly lengthens the air tube, the student notices no more resonance until the tube is $0.487meters$ longer than the previous resonant length. This is supposed. From this information, the student calculates the speed of sound. What value does he calculate?
A) $84m/s$
B) $342m/s$
C) $168m/s$
D) $351m/s$
E) $336m/s$

Answer 1

Hint: The standing waves in the open tube air column produces the desired frequency that replicates the same frequency that the tuning for produces. You can understand that the resonance effect clearly signifies this fact. The specific length is the resonant length. After lengthening the tube, it again resonates with the same frequency. This supposed to be the second overtone of the open pipe.

Formula used:
The frequency $f$ of the ${n^{th}}$ harmonics of the vibration of the open tube is
$f = \dfrac{{n\nu }}{{2L}}$ ……….(1)
where L is the length of the open tube and $v$ is the velocity of the sound.

Complete step by step answer:
Given:
The frequency of the tuning fork and also the mentioned resonant frequency is $345Hz$.
The open pipe by extended by $0.487meters$.
To get: Velocity of sound $v$.

Step 1:
Let the resonant length of the open tube is $l$.
Now, calculate the frequency of the first harmonic of the open tube air column
$f = \dfrac{{1 \times v}}{{2l}}$
This is the same as the resonant frequency of $345Hz$.
$345 = \dfrac{v}{{2l}}$ ……………….(2)

Step 2:
When the tube is lengthened it becomes $\left( {l + 0.487} \right)meters$.
Now calculate the frequency of the second harmonic of the air tube from eq (1).
$f{'} = \dfrac{{2 \times v}}{{2 \times \left( {l + 0.487} \right)}} $
This again is equal to the same resonant frequency.
$\Rightarrow 345 = \dfrac{{2 \times v}}{{2 \times \left( {l + 0.487} \right)}} = \dfrac{v}{{l + 0.487}}$

Step 3:
Now, compare the two frequencies.
$\dfrac{v}{{2l}} = \dfrac{v}{{l + 0.487}}$
$ \Rightarrow l = 0.487meters$

Step 4:
Now put this back to eq (2).
$ 345 = \dfrac{v}{{2 \times 0.487}} $
On simplification,
$ \Rightarrow v = 345 \times 2 \times 0.487 $
$\therefore v = 336.03m/s$

The student calculates the velocity of sound as $336m/s$. Hence, option (E) is correct.

Note:
Here you may make mistakes in understanding that the resonance happens when the frequencies of the tuning fork and the frequency of the open tube matches. Again, you should note that after lengthening, the resonant frequency occurs at the second harmonics of the open tube.