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Calculation of the period and the spring constant by observing the harmonic motion of a mass attached to the end of a spring, ...
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Objective:
Calculation of the period and the spring constant by observing the harmonic motion of a mass
attached to the end of a spring, the weight of the mass connected to the springs and the spring
constant according to the oscillation period
Materials:
Spring, ruler, timer, masses of 100, 200 and phet simulation
(https://phet.colorado.edu/sims/html/masses-and-springs/latest/masses-and-springs_en.html)
Theoretical background:
Consider a spring that extended some distance by a force of F, in this case the relationship
between the restoring force F and the displacement from equilibrium position x is given as
follows
Here F is the force on the spring, x is the displacement of the end of the spring from its
equilibrium position, and k is the spring constant, a property of the spring. For springs this
relation is known as Hookeโs Law. The weight of a mass suspended to the end of a spring is the
stretching force of the spring. Hanging different weights to the end of a spring and measuring the
extensions allow us to test the springโs elasticity. The value of k is a measure of stiffness. A stiff
spring will have a large value for k.
Suppose you hang a spring with force constant k and suspend from it a body with mass m. Now
m is at equilibrium position, x. If you pull the mass down little bit and let it go m start moving
around x position and its harmonic motion is observed. a simple harmonic motion, the force
acting on the object is proportional to the distance of the object for equilibrium. The elapsed time
of the mass m passing two times from any fixed point is called the Oscillation Period of
harmonic motion. The oscillation period T depends on the spring constant k and the mass, m, of
attached body and it is given by
๐ = 2๐โ ๐/k
Procedure:
record this elongation (displacement) as x= ฮ l
acceleration; 9.8 m/s
2
and divided this elongation distance to find k spring
constant value and recorded it.
counted 10 full turns.
the k again by using T= 2๐โ ๐/k
M=100 gr K from f = k. โ l K from T= 2๐โ ๐/k
0.1 kg 6.1 N/m 5.9 N/m
Table 1
M=200 gr K from f = k. โ l K from T= 2๐โ ๐/k
0.2 kg 6.3 6.
Table 2
f = k. โ l
Calculations:
i
=48cm = 0.48m
f
= 64cm = 0.64m
โ l = 0.64-0.48 = 0.16 m
K= f/
โ l f = mg = 0.1 x 9.81 = 0.981 N
0.981/0.16 = 6.1 N/m.
T = 0.812 s => 10T = 8.12s
T = 2 ฯ
m
k
2
ฯ
2 m
k
To fin k :
k = 4 ฯ
2
m
2