Spring constant
APPARATUS:
1.
2.
3.
4.
5.
6.
Spring
Rigid support
Slotted weights
Vertical wooden scale
Fine pointer
Hook.
As
Theory:
A spring is an elastic object which stores mechanical energy. Springs are usually made of
hardened steel. The spring constant, k, of an ideal spring is defined as the force per unit length
and is different from one spring to another. Spring constant is represented in Newton/meter
(N/m). It can be determined both in static (motionless) as well as dynamic (in motion)
conditions. Two different techniques are used for determination of the spring constant. In the
static method, Newton’s second Law is used for the equilibrium case, and laws of periodic
motion are applied for determining the spring constant in the dynamic case.
Static Mode or Hooke’s Law:
The restoring force, F, of a stretched spring is proportional to its elongation, x, if the
deformation is not too great. When a load F suspended from lower free end of a spring
hanging from a rigid support, it increases its length by amount x, then
Fα-x
F=-kx
K=-F/x
where k is constant of proportionality. It is called the force constant or the spring constant of
the spring. The spring’s restoring force acts in the opposite direction to its elongation,
denoted by the negative sign.
Procedure:



Read the position of the pointer on the scale. Avoid the parallax error. - This is done
by reading the position of the pointer on the scale when the image of the pointer is
hidden behind the pointer. (Remember the procedure of the experiment on
equilibrium.)
Increase the load on the spring in equal steps and read the position of the pointer each
time. Thus, fill out the first two columns of the first table of the data sheet.
Now gently pull the load down through about Q.5 cm. Release it gently and let it come
to rest. Take the reading of the position of the pointer and enter it in the last row of
column 3 of the table. Gently decrease the load in equal steps and thus fill out the rest
Diagram:
Spring System (weight Attached)
Calculations:
No of
observations
Load
F = mg
1
2
3
4
5
490,000
980,000
1470000
1960000
2450000
Load
Load
Increased decreased
25
26
30
34
39
25.1
26
30
34
39.2
Mean
(X2)
25.05
26
30
34
39.1
X=
X2 - X1
0.9
1.9
5.9
9.9
13.9
K=
𝐟
𝐱
544444.5
515789.5
249152.5
197979.5
176258.9
Mean K = 336724.9 Dynes / cen
Dynamic Mode:
If a mass M is suspended from a spring of spring constant k and the system is made to oscillate,
then for small amplitudes of oscillations along the length of the spring, the motion of the
system is simple harmonic. In this case, the time period T of the system is given by
𝐌+𝐌𝐩
T = 2𝛑√
𝐤
Eq. 1
where Mp is the effective mass of the spring which is approximately equal to 1 /3 the mass of
the spring.
Eq. (1) gives
T2 =
𝟒𝝅𝟐
𝒌
M+
𝟒𝝅𝟐
𝒌
Mp
Thus, if we plot T2 vs. M, then for a given spring (k and MP constant), the graph will be a
straight line. By choosing two points (M1, T12) and (M2,T22) on the graph, we get
T 12 =
𝟒𝝅𝟐
𝒌
M1 +
𝟒𝝅𝟐
𝒌
Mp
Similarly
T 22 =
𝟒𝝅𝟐
𝒌
M2 +
𝟒𝝅𝟐
𝒌
Mp
Subtracting (1) From (2)
K = 4π2
𝑴𝟐−𝑴𝟏
𝑻𝟐𝟐 − 𝑻𝟐𝟏
𝐦
K = 4π2 𝐓𝟐
Procedure:
1.
2.
3.
4.
5.
Place a suitable load on the hanger
start the oscillations of the system.
Make sure that the oscillations are along the length of the spring and the amplitude is small.
Find the time for 40 (or 50) - oscillations 3 times.
Repeat the procedure by changing the load on the spring in equal steps.
Calculations:
No of vibrations taken = 20
No of
observation
Mass
T1
T2
Mean
(t)
1
2
3
4
5
1000
1500
2000
2500
3000
11.09
16
17
23
26.4
11.03
16.75
17.75
23.4
26
11.06
16.3
17.4
23.2
26.2
𝐦
As K = 4π2 𝐓𝟐
Here m is mean of masses = 2000
K = 4 π2 x 2000/1.424
= 554472.1
T=
𝐭
𝟐𝟎
0.55
0.81
0.87
1.16
1.31
T2
0.3
0.656
1.012
1.368
1.789