Question

In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form?

Answer 1

Explanation:

(a) Given that frequency ν = 400 Hz. The length of air column for the fundamental frequency of resonance =L₁ = 20.0 cm. The length of air column for the first overtone frequency of resonance =L₂ = 62.0 cm. If the pressure node forms at d above the open end, Hence (L₂+d) - (L₁+d) = /2 {See the diagram below} →62.0 - 20.0 cm =/2 → = 2*42.0 cm =84.0 cm =0.84 m Hence the speed of sound in air V = ν =400*0.84 m/s =336 m/s (b) Since L₁+d =/4 {From figure for fundamental vibration} Hence, d = /4-L₁ =84/4 -20 =21 -20 cm = 1.0 cm.Read more on Sarthaks.com - https://www.sarthaks.com/325604/resonance-column-experiment-tuning-frequency-first-resonance-observed-when-column-length

Answer 2

(a) The speed of sound in air is $0.84 \mathrm{m}$.

(b) The distance above the open end does the pressure node form is $1.0 \mathrm{cm}$

Explanation:

(a) Given that frequency ν = 400 Hz.  

The length of the air column to the fundamental resonance frequency =L₁ = 20.0 cm.  

Duration of the air column for first resonance overtone frequency =L₂ = 62.0 cm.  

If the pressure node above the open end is in d,

Hence $\left(L_{2}+d\right)-\left(L_{1}+d\right)=\frac{\lambda}{2}$ {See the diagram below}  

$62.0-20.0 \mathrm{cm}=\frac{\lambda}{2}$

$2^{*} 42.0 \mathrm{cm}=84.0 \mathrm{cm}=0.84 \mathrm{m}$

The speed of sound in the air is therefore $\mathrm{V}=\mathrm{v}=400^{*} 0.84 \mathrm{m} / \mathrm{s}=336 \mathrm{m} / \mathrm{s}$

(b) Since $\mathrm{L}_{1}+\mathrm{d}=\frac{\lambda}{4}$ { From the Simple Vibration calculation }  

Hence, $\mathrm{d}=\frac{\lambda}{4}-\mathrm{L}_{1}=\frac{84}{4}-20=21-20 \mathrm{cm}=1.0 \mathrm{cm}$