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Resonance Column Experiment (Class 11) Readings

 

Resonance Column Experiment (Class 11) Readings

Experiment – 1


Aim


i. To find the velocity of sound in air at room temperature and hence at 0°C using a resonance column apparatus.

ii. To find the unknown frequency of the given tuning fork.


Apparatus


Resonance column apparatus, Tuning forks, Rubber hammer, Meter Scale etc.


Principle:


i. The velocity of sound at room temperature, Vt = 2n(l2-l1)

where, n = Frequency of the tuning fork

l1 = First resonating length

l2 = Second resonating length

ii. The velocity of sound at 0°C is given by V0 = Vt(273/(273+t))

or V0 = Vt – 0.6t

where, t = Room temperature°C

iii. Unknown frequency, n’ = Vt/(2(l2’-l1’))


Procedure


Length of the air column is kept very small. A tuning fork of known frequency (n) is excited and held horizontally over the mouth of the inner tube. The length of air column in the inner tube is slowly increased by raising the tube till a booming sound is heard. The length of the air column is measured (l1). Keeping the tuning fork excited at the mouth of the tube the length of the air column is measured (l1). Keeping the tuning fork excited at the mouth of the tube the length of the air column is increased further. The length (l2) of the air column is measured when the booming sound is heard (l2>3l1). The experiment is repeated and the mean values of l1 and l2 are found out. The velocity of sound at room temperature is calculated. The experiment is repeated for different tuning forks and the mean value of Vt is found. From this the velocity of sound at 0°C is calculated.


Using the tuning forks of known frequencies the mean value of Vt is calculated. Then using the tuning fork of unknown frequency, the first and second resonating lengths (l1’ and l2’) are measured. The unknown frequency (n’) can be calculated.


Observations and Readings


Frequency of tuning fork (n)

First resonance length (l1)

Second resonance length (l2)

Vt

1(cm)

2(cm)

Mean (l1’)

1(cm)

2(cm)

Mean (l2’)

512

16.5

16.5

16.5

50.5

50.5

50.5

34816

480

17.5

17.5

17.5

53.5

53.5

53.5

34560

Unknown

22.5

22.5

22.5

68.5

68.5

68.5

 

 

Mean Vt = 34688 cm/s = 346.88 m/s

Room temperature, t = 30°C

Velocity of sound at 0°C, Vo = Vt – 0.6t = 346.88 – 0.6 x 30 = 328.88 m/s

Unknown Frequency, n’ = Vt/(2(l2’-l1’)) = 403.3 Hz


Results:


1. Velocity of sound at room temperature = 346.88 m/s

2. Velocity of sound at 0°C = 328.88 m/s

3. Unknown frequency = 403.3 Hz


Experiment - 2


Aim: To compare the frequencies of two tuning forks and also to determine the end correction.


Principle:


Let l1 and l2 are the first and second resonance length with a tuning fork of frequency n1, l1’ and l2’ respectively are the first and second resonance length with another tuning fork of frequency n2. Then,


i. Ratio of frequencies, n1/n2 = (l2’-l1’)/(l2-l1)

ii. The end correction is given by, e = (l2-3l1)/2


Procedure:


The length of the air column is kept very small. The first tuning fork of frequency n1 is excited and is held horizontally close to the mouth of the inner tube. The inner tube is slowly raised until maximum sound is heard. The length of air column is measured as l1. Then the inner tube is further raised, keeping the vibrating fork at the mouth of the tube, till the maximum sound is heard. The length of air column noted as l2. Repeating this procedure for another tuning fork of frequency n2 and the resonance lengths are measured as l1’ and l2’. From this n1:n2 is found out. The end correction is also calculated.


Observations and Readings


1. To compare the frequencies


Frequency of tuning fork (n)

First resonance length

Second resonance length

(l2’-l1’) /

(l2-l1)

1(cm)

2(cm)

Mean

1(cm)

2(cm)

Mean

n1 = 512

16.5

16.5

l1 = 16.5

50.5

50.5

l2 = 50.5

1.06

n2 = 480

17.5

17.5

l1’ = 17.5

53.5

53.5

l2’ = 53.5

 

Ratio of frequencies, n1/n2 = 1.06

(l2’-l1’)/(l2-l1) = 1.06


2. To find the end correction


Tuning fork

First resonance length (l1)

Second resonance length (l2)

e

1(cm)

2(cm)

Mean

1(cm)

2(cm)

Mean

First

16.5

16.5

16.5

50.5

50.5

50.5

0.50

Second

17.5

17.5

17.5

53.5

53.5

53.5

0.50

 

Mean e = 0.5 cm = 0.5 x 10-2 m


Results:


i. Frequencies of two tuning forks are compared.

ii. End correction = 0.5 x 10-2 m

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