Question

A long glass tube is held vertically in the water. A tuning fork is struck and held over the tube. Strong resonance is observed at two successive lengths of $0.5\text{m and 0}\text{.84m}$ above the surface of the water. If the velocity of sound is $340\text{ m/s}$, then the frequency of the sound is (neglect end corrections)
A. $128Hz$
B. $256Hz$
C. $384Hz$
D. $500Hz$

Answer 1

Hint: The glass tube held above water will act as an organ pipe which is open at one end. The water surface will act as the closed surface while the other end will act as the open end. The difference between the two resonant lengths will be equal to the difference between the two consecutive conditions for resonance in an organ pipe.

Complete step by step answer:
As the glass tube is moved up and down the water, the resonance is occurring at two successive lengths of $0.5\text{m and 0}\text{.84m}$respectively. The velocity of sound in air is given $340\text{ m/s}$. The glass tube can be considered as an organ pipe with one of its ends closed. The closed surface, in this case, is the surface of the water.
The condition for resonance in an organ pipe which is closed at one end and open at the other is given by,
$l=\left( 2n+1 \right)\dfrac{\lambda }{4}$
Where
n is the mode of resonance.
l is the resonating length of the glass tube.
$\lambda$ is the wavelength of sound in the glass tube.
Here two successive resonances occur at lengths $0.5\text{m and 0}\text{.84m}$. So, we can write,
$0.5m=\left( 2n+1 \right)\dfrac{\lambda }{4}$ …. Equation (1)
$0.84m=\left( 2\left( n+1 \right)+1 \right)\dfrac{\lambda }{4}$ …. Equation (2)
Taking the difference between Equation (2) and Equation (1), we get,
$0.84m-0.5m=\dfrac{\lambda }{2}$
$\Rightarrow \lambda =2\left( 0.84m-0.5m \right)$
$\therefore \lambda =0.68m$
We now know the wavelength of the sound wave produced by the tuning fork. So, the frequency of the sound wave can be found out by using the relation,
$\text{v}=\upsilon \lambda$
Where,
v is the velocity of the sound wave.
$\lambda$ is the wavelength of the sound wave produced by the tuning fork.
$\upsilon$ is the frequency of a sound wave produced by the tuning fork.
Substituting the values in the above Equation, we get,
$\upsilon =\dfrac{340m/s}{0.68m}$
$\therefore \upsilon =500Hz$
So, the frequency of the tuning fork is 500 Hz.
So, the answer to the question is option (D).
Note:
End correction is necessary when we are dealing with an organ pipe, and there is a chance of error in the formation of the wave because the wave may not actually start from the opening of the object but instead can start before the opening, thus resulting in an error. Hence an end correction is sometimes required to appropriately study its properties. The end correction depends on the radius of the diameter of the object.
In end correction, a short distance is added to the original length of the organ pipe in order to remove the error. So, in reality, the actual pitch of an organ pipe will be less than the theoretical pitch of the pipe. The end correction is usually denoted by the letter ‘e’.