## Random Posts ## 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.

 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.

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 ## Helical Spring Experiment (Class 11) Readings

Helical Spring Experiment with Readings (Class 11)

Experiment - 1

Aim: To find the spring constant k of a helical spring by measuring time period of vertical oscillations of a known load. ## Simple Pendulum Experiment (Class 11) Readings

Simple Pendulum Experiment (Class 11) with Calculations and Readings

Aim

i. To verify the relation between period (T) and length (l) of pendulum

ii. To find out the acceleration due to gravity

iii. To determine the length of seconds pendulum

iv. To find the period of the pendulum whose length is 105 cm.

Apparatus

Simple pendulum, Stop-watch, Meter scale, Wooden block etc..

Principle

For small amplitudes of oscillation of a simple pendulum,

l/T2 = a constant where, l —> Length of the pendulum

T—> Period of the pendulum

Also, T = 2π(l/g) where, g —> Acceleration due to gravity at the place

Therefore, g = 4π2(1/T2)

Procedure

i. Relation between I and T2

The bob is placed between two wooden blocks and its diameter (d) is measured. Hence radius (r = d/2) can be calculated. The length of the string (I) is so adjusted that the distance between the point of suspension and the bottom of the bob is (50 + r) cm. Hence the length of the pendulum ‘l’ is 50 cm.

A mark is made on the edge of the table using a piece of chalk to indicate the equilibrium position of the pendulum, ie. , the position of the pendulum when it is at rest. The bob is pulled aside through a small distance and released. The pendulum executes oscillations, after a few oscillations, a stop watch is started just when the pendulum crosses the equilibrium position. When it next crosses the mark in the same direction, the oscillations are counted and when the pendulum passes the chalk mark in the same direction for the 20th time, the stop watch is stopped.

Thus the time taken for 20 oscillations is determined. This is repeated and the mean time taken for 20 oscillations is found. Hence the time for one oscillation ie., period 'T' is calculated. l/T2 is also calculated.

The experiment is repeated for lengths 60, 70, 80, 90, 100,110 cm. In each case l/T2 is calculated. It is found to be a constant. This is the relation between the period and length of a simple pendulum. Graph is drawn with 'l' along the X-axis and 'T2' along the Y-axis. A straight line graph is obtained.

ii. Acceleration due to gravity at the place

The mean value of l/T2 is determined. Hence 'g' can be calculated using the formula g = 4π2(1/T2). Note AB and BC from the graph. Hence g can be calculated using the equation, g = 4π2.AB/BC

iii. Length of the seconds pendulum

For a seconds pendulum, T = 2 seconds. Then T2= 4. From the graph, the value of 'l’ for which T2 = 4 is found (OD). This gives the length of the seconds pendulum.

iv. Period of the pendulum whose length is 105 cm

From the graph the square of the period corresponding to the length 105 cm is noted (OE). The square root of this ( OE ) gives the period.

Radius of the bob, r = 0.95 cm

 No: Distance between point of suspension and the bottom of the bob (l+r) Length of the pendulum (l) Time for 20 oscillations (t) Period of oscillation (T=t/20) T2 l/T2 1 2 mean cm cm s s s s s2 cm/s2 1 2 3 4 5 6 7 50.95 60.95 70.95 80.95 90.95 100.95 110.95 50 60 70 80 90 100 110 28 30 31 34 35 38 40 28 30 31 34 35 38 40 28 30 31 34 35 38 40 1.4 1.5 1.55 1.7 1.75 1.9 2 1.96 2.25 2.40 2.89 3.06 3.61 4 25.5 26.7 29.7 27.7 29.4 27.7 27.5

Mean value of 1/T2 = 27.661 cm/s2 = 0.276 m/s2

Therefore, Acceleration due to gravity, g = 4π2(1/T2) = 4 x (3.14)2 x 0.276 = 10.896 m/s2

l – T2 graph

From graph:

AB = 43 cm = 0.43 m

BC = 1.5 s2

G = 4π2(AB/BC) = 4 x (3.14)2 x 0.43/1.5 = 11.30 m/s2

Length of the seconds pendulum (OD) = 110 cm = 1.10 m

Period of the pendulum whose length is 105 cm = OE = 1.92 s

Results

i. l/T2 is found to be a constant

ii. Acceleration due to gravity at the place,

(a) by calculation = 10.896 m/s2

(b) from graph = 11.30 m/s2

iii. Length of the seconds pendulum =1.10 m

iv. Period of the pendulum whose length is 105 cm = 1.92 s 