Question

In a resonance column experiment, a tuning fork of frequency 400Hz is used. The first resonance is observed when the air column has a length of 20.0cm and the second resonance is observed when the air column has a length of 62.0cm.
a) Find the speed of sound in the air.
b) How much distance above the open end does the pressure node form?
A. 336,1
B. 400,3
C. 500,5
D. 600,0

Answer 1

Hint: We will solve the first part by finding out the wavelength by subtracting the lengths of the air column given in the question. Then with the formula of frequency, we will find out the velocity of the sound. Refer to the question for further queries.

Complete Step-by-Step solution:



a) The given frequency of the tuning fork as per given in the question is- 400Hz
The length of the air column when the first resonance is observed- L1 = 20.0cm
The length of the air column when the second resonance is observed- L2 = 62.0cm
We will now find the wavelength using the formula used in the condition of resonance i.e. $\lambda = 2\left( {{L_2} - {L_1}} \right)$.
Putting the values of L1 and L2 in the formula mentioned above, we get-
$
   \Rightarrow \lambda = 2\left( {62 - 20} \right) \\
    \\
   \Rightarrow \lambda = 2 \times 42 \\
    \\
   \Rightarrow \lambda = 84cm = 0.84m \\
$
Since we know that- $f = \dfrac{v}{\lambda }$, putting the values we get the value of the velocity i.e.-
$
   \Rightarrow f = \dfrac{v}{\lambda } \\
    \\
   \Rightarrow v = f \times \lambda \\
    \\
   \Rightarrow v = 400 \times 0.84 \\
    \\
   \Rightarrow v = 336m/s \\
$

b) Let the distance at which the pressure node forms be d.
So, the total length from the end of the air column to the distance d will be – L1+d. The given distance is equal to $\dfrac{\lambda }{4}$ (since the distance between node and the anti-node is $\dfrac{\lambda }{4}$). Thus,
$
   \Rightarrow {L_1} + d = \dfrac{\lambda }{4} \\
    \\
   \Rightarrow d = \dfrac{{84}}{4} - 20 \\
    \\
   \Rightarrow d = 21 - 20 \\
    \\
   \Rightarrow d = 1cm \\
$
Hence, the pressure node forms 1cm above the open end.
Thus, option A is the correct option.

Note: The distance between the node and the anti-node is always considered to be $\dfrac{\lambda }{4}$. In the above question, the open end and the point at which the pressure node formed were considered as the node and the anti-node.