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3B Scientific Oscillation Tube User Manual

Page 5

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5

p V = n R T

(1)

For changes in state where no exchange of heat with the
environment takes place, this equation can be reduced
to the adiabatic formula:

p V

?

= const.

(2)

The adiabatic exponent

? is the ratio of the specific heat

at constant pressure c

p

to the specific heat at constant

volume c

V

:

χ =

c

c

V

P

(3)

A stopper with a hole is put into a glass vessel with a
volume of 10 l . The precision tube runs through the hole
in this stopper so that it is in a vertical position. If the
aluminium cylinder is allowed to drop into the tube, it
bounces on the cushion of air enclosed inside the appa-
ratus. This leads to a periodic oscillation.
When the pressure p in the glass vessel is equal to the
sum of the pressure due to the mass of the aluminium
cylinder and the external atmospheric pressure, the cyl-
inder is in a state of equilibrium:

p

p

mg

A

= +

L

(4)

If the cylinder is moved a distance s from its equilibrium
position, p changes by the value

,p and V changes by

,V. A force acts on the aluminium cylinder to push it
back towards equilibrium. This force is proportional to
the distance s. A harmonic oscillation now begins atop
the air cushion under the cylinder. Since the oscillation
occurs fairly rapidly, it can be described in terms of the
adiabatic change of state. By deriving dp/dV from equa-
tion (2) and assuming this also applies to the small finite
changes

,p and ,V we obtain

p

p

V

V

= −

χ

(5)

Since the cylinder moves along a distance s in the preci-
sion tube, the change in volume is

,V = As

(6)

The restoring force

F

A p

pA

V

s

=

= −

χ

2

(7)

leads to the periodic acceleration of a cylinder of mass m.
Newton’s second law then gives us the following differ-
ential equation for s(t)

d s

dt

pA

V

s

2

2

2

+

=

χ

0

(8)

From (8) the natural frequency of oscillation is given by

ω

χ

=

pA

V

2

(9)

thus the period of the oscillation T

s

is

T

mV

pA

s

2

2

2

=

=

π

ω

π

χ

(10)

Therefore, to derive the adiabatic exponent

? the follow-

ing applies:

χ

π

=

=

4

64

2

2

2

2

4

mV

A pT

mV

T d p

s

s

(11)

4. Operation

• Determine the atmospheric pressure, the diameter

of the inside of the precision tube, the mass of the
aluminium cylinder and the volume of the measur-
ing flask.

• Place the glass tube on the Marriott bottle, make sure

it is vertical and secure it to a stand.

• The Marriott bottle should have a rubber mat or simi-

lar placed inside to avoid damage to the bottle and
the cylinder if the cylinder falls into the bottle.

• To simplify the experiment, it is recommended that a

hand pump be attached to the Marriott bottle via the
3-way stopcock. The cylinder can then be pumped
back up the tube and retrieved from the top so that it
is not necessary to repeatedly set the tube up.

• The aluminium cylinder should be cleaned with a

fluff-free cloth and some petroleum ether. It should
not be introduced into the tube at a crooked angle
and allowed to drop when the stopcock is closed. Do
not touch the cylinder except by the grip to prevent it
getting dirty.

• Measure the duration of 5 oscillations 10 times using

a stopwatch.

• The time measurement should start at the point when

the cylinder comes to a halt for the first time at its
lowest point. The watch should be stopped when the
cylinder reaches its lowest point for the sixth time.

• Using the hand pump with the stopcock open, pump

the cylinder back to the top. Make sure that the cylin-
der does not fall out at the end and get damaged.

• Take the cylinder right out of the tube so that the

pressure in the apparatus returns to atmospheric pres-
sure. Close the stopcock again.

• Make the measurements another nine times and

determine the average value for the time.

• Perform the calculation.