Question

A piston is fitted in a cylindrical tube of small cross section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the speed of sound in the medium of the tube.

• A. $328\frac{m}{s}$
• B. $512\frac{m}{s}$
• C. $448\frac{m}{s}$
• D. None of these

Answer 1

The correct option is A

$328\frac{m}{s}$

So if you have understood the question you will realize that we are talking above closed tube. One end open while the other end is closed by piston itself.

So we can change the length of the tube in which standing wave is formed by moving the piston. Correct!!!
Now that you have identified the system, lets try to recall about the concepts and formulas being used.
Before you write down any formula's and start calculating its always better to draw diagram and visualize what is happening
So clearly in an open pipe the difference between 2 consecutive lengths that support resonance is

Here in our question

Also $\lambda =\frac{v}{f}$

$V=2f\times 32\times {10}^{-2}$ $\frac{\lambda }{2}=32cm=32\times {10}^{-2}m$

frequency is given to be 512 Hz as frequency of sound depends on the source which here is a tuning fork.

$\Rightarrow$ $V=2f\times 32\times {10}^{-2}$

$V=328\frac{m}{s}$