Question

A uniform tube 60 cm long, stands vertically with lower end dipping into water. When its length above water is 16 cm and again when its is 48 cm, the tube resonates to a vibrating tuning fork of frequency 512 Hz. The lowest frequency to which this tube can resonate when it is taken out of water is nearly
A. 819 Hz
B. 773 Hz
C. 273 Hz
D. 256 Hz

Answer 1

The nth harmonic (frequency of a standing wave) in an air column with one end closed is given as ${{f}_{n}}=\dfrac{nv}{4l}$, where v is the speed of the wave and l is the length of the air column.
The nth harmonic in an air column with both ends open is given as ${{f}_{n}}=\dfrac{nv}{2l}$.

Formula used:
${{f}_{n}}=\dfrac{nv}{4l}$
${{f}_{n}}=\dfrac{nv}{2l}$
where $v$ is the speed of the wave and $l$ is the length of the air column

Complete step by step answer:
The nth harmonic (frequency of a standing wave) in an air column with one end closed is given as ${{f}_{n}}=\dfrac{nv}{4l}$, where v is the speed of the wave and l is the length of the air column. In this case, the tube is dipped in water and it is given that the tube resonates with a vibrating tuning fork of frequency 512 Hz for the first time at 16 cm length of the air column. This means that the frequency of the standing wave in the air column is equal to 512 Hz.Therefore, ${{f}_{n}}=512Hz$ and $l=16cm=0.16m$.
Let this frequency be ${{n}_{1}}^{th}$ harmonic.
$\Rightarrow 512=\dfrac{{{n}_{1}}v}{4(0.16)}$ …… (i).
Then the tube resonates with the same frequency for the second time when the length of the air column is 48 cm. Let this frequency be ${{n}_{2}}^{th}$ harmonic.
$\Rightarrow 512=\dfrac{{{n}_{2}}v}{4(0.48)}$ ….. (ii).
Now, divide (i) by (ii).
$\Rightarrow \dfrac{512}{512}=\dfrac{\dfrac{{{n}_{1}}v}{4(0.16)}}{\dfrac{{{n}_{2}}v}{4(0.48)}}$
$\Rightarrow 1=\dfrac{{{n}_{1}}(0.48)}{{{n}_{2}}(0.16)}$
$\Rightarrow \dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{1}{3}$.
The value of n can be a natural number only and we cannot have any standing wave for length greater than 48cm.
This means that ${{n}_{1}}=1$ and ${{n}_{2}}=3$.
Substitute the value of ${{n}_{1}}$ in (i).
$\Rightarrow 512=\dfrac{(1)v}{4(0.16)}$
$\Rightarrow v=512\times 4\times 0.16=327.68m{{s}^{-1}}$.
This means that the speed of the sound in that medium is $327.68m{{s}^{-1}}$.
The nth harmonic (frequency of a standing wave) in an air column with both ends open is given as,
${{f}_{n}}=\dfrac{nv}{2l}$ …. (iii).
The lowest resonating frequency is the first harmonic. For first harmonic $n=1$.
We found that $v=327.68m{{s}^{-1}}$ and the length of the air column is $l=60cm=0.6m$.
Substitute the values in (iii).
$\therefore{{f}_{n}}=\dfrac{(1)(327.68)}{2(0.6)}=273.06Hz$
Therefore, the lowest frequency to which this tube can resonate when it is taken out of water is nearly 273 Hz.

Hence, the correct option is C.

Note: Note that there are two different cases where a standing wave can form.

1. Both the ends of the tube open: In this, the harmonics are called first harmonic, second harmonic and so on. The value of n is a natural number.
2. Only one end is closed: In this, the harmonics are called first harmonic, third harmonic, fifth harmonic and so on. The value of n is an odd natural number.