A 50 cm long hollow glass cylinder is open at both ends and is suspended in air.

Question

A 50 cm long hollow glass cylinder is open at both ends and is suspended in air. A source of sound that produces a pure frequency is placed close to one end of the tube. The frequency of the sound source is slowly increased from a very low value. The tube is first observed to resonate when the source frequency is 320 Hz.

(i) The end of the cylinder opposite the end where the sound source is positioned is tightly sealed with a rigid metal plate. Calculate the first THREE resonant frequencies (lowest to highest).

(ii) If the apparatus is now placed in helium gas instead of air, do the resonant frequencies referred to in previous parts of this question increase or decrease? Give your reasoning.

(iii) The original air-filled apparatus now suffers a temp decrease of 40oC, do the resonant frequencies referred to in previous parts of this question increase or decrease? Give your reasoning.

Explanation / Answer

1.>from the given condition:
when both ends are open
1st resonance occurs at 320 Hz

/2=length of the column =0.5 m

  =1m

velocity of sound ,v=320*1=320m/s

now ,when one end of the column is closed

for resonance

length of column=l=(2n-1)/4 for n=1,2,3,4....

frequency,f=velocity of sound /

for n=1   =4l=2m

frequency,f1=320/2 =160Hz

for n=2 =4l/3=2/3 m

f2= 320*3/2 =480 Hz

for n=3 =4l/5=0.4 m

f3=320/0.4 =800Hz

hence three resonant frequencies are 160 Hz,480Hz and 800 Hz

3.>with decrease in temperature speed of sound in air also decreases and frequency is directly proportional to the speed of sound

all the three frequencies will decrease from their original value

2.>in helium the speed of sound increases so the values of resonant frequencies will also increase from their original values

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