Question

In a standing wave pattern in a vibrating air column, nodes are formed at a distance of 4.0 cm. If the speed of sound in air is 328 m s−1, what is the frequency of the source?

Answer 1

Answer:

.....................

Answer 2

The frequency of the source is $4.10 \mathrm{kHz}$

Explanation:

Step 1:  

Given data,  

The distance between the two consecutive nodes is comparitevely equal to the half of the vibration's wavelength.

This distance has been given here  = 4.0 cm

Converting centimeter to meter  

$\frac{4}{100}=0.04 \mathrm{m}$

$\text { Hence } \frac{\lambda}{2}=0.04 \mathrm{m}$

$\lambda=2^{*} 0.04 \mathrm{m}=0.08 \mathrm{m}$

Step 2:

Because sound speeds in the air,

$\mathrm{V}=328 \mathrm{m} / \mathrm{s}$

So the frequency of the source,  

$f=\frac{v}{\lambda} f=\frac{328}{0.08} \mathrm{Hz}$

$f=\frac{32800}{8} H z$

$f=4100 \mathrm{Hz}$

$=4.10 \mathrm{kHz}$