of Elastic Constant of Spiral Spring, and Earth's Gravitational Intensity
. Elastic Constant Of Spring
idea is to investigate the relationship for the extension of spiral spring x and for the oscillation period of this spring T of suspended different mass ma mass is suspended from the end of a spring, Hooke's Law states that that the extension of the spring is proportional to the mass provided the elastic limit of the spring is not exceeded., the tension force, F (N), in the spring is proportional to the extension x (m) produced. That is:
=kx ,
k (N/m) is the spring (elastic) constant.
a mass m is placed on the spring, acting on the mass gravitational field of the Earth. Then the spring is stretched until the force of gravity mg is balanced by the elastic force of the spring F (Fig.1). And given by
=mg=kx. (1)
the tension of the spring is then mg., by measuring the mass of the load m and the corresponding extension x (OB, Fig.1) of the spring, it is possible to find the ratio the acceleration due to gravity at this point on the Earth's surface by the spring constant this spiral spring (k/g) .equation (1):
, (2)
:=m + m 0 - total mass of load on spring, which stretches a spring (kg), m - the total mass of load (kg), m 0 - the mass of holder (kg ); - extension of spring, mean average reading length spring minus reading initial length spring x=l - l 0) .k/g - the gradient a graph of extension x against the mass M (kg/m)., we can plot a graph of extension x against the mass M whose weight extended the spring to found the ratio the acceleration due to gravity at this point on the Earth's surface by the spring constant this spiral spring (k/g) .this we must draw the best straight line through the origin. From the graph calculate the gradient=(k/g) .Variable: mass of load on spring (m) .Variable: length spring after stretches (l) .Variables: initial length spring l 0; mass of holder m 0.
. Earth Gravitational Intensity (g)
the mass is pulled down a little and then released, it oscillates up and down above and below O (position of equilibrium) (Fig1, b) .true the motion is simple harmonic about O, and the period oscillation of spring is given by:
. (3)
T - period of oscillation of the spring (s); - total mass of load on spring (kg); - where m is a constant depending on the mass of the spring itself; - elastic constant ( N/m). (3), it follows that:
) =. (4)
of oscillation of the spring T we found in this way:
=t/N,
: - time taken N oscillation (s); - number of oscillation.a graph of T 2 against M should be a straight line and the y-intercept is m.of this graph gives us:
=.
we can found elastic constant k for this spiral spring.the value of acceleration due to gravity at this point on the Earth's surface can be found by substituting the magnitudes of gradient at part 1 and elastic constant k in part 2:
g=k/gradient.
data:
Zero reading of the spring (initial length of the spring) l 0=0,02 m.holders B to the spring m 0=0,049 kg.
Error:
To measure the initial length and length of the spring after extension I used a ruler, the absolute uncertainty of which was equal to ± 0,5 mm, because the limit uncertainty metre rule is half the limit of reading (1 mm) .of holder was measured on a digital balance with a precision of ± 0.1 gmass of load was measured on a digital balance with a precision of 0.1 g. In all cases the measured mass was less than 1 g off. A typical measure is m1=199.3 g. The masses are thus assume d to be accurate to ± 1 g or? M=± 0.001kg. measure the time of 10 complete oscillations I used stopwatch, the absolute uncertainty of which was equal to ± 0.1 s.as to obtain more accurate data, I measured time of 10 complete oscillations by load decreasing for same mass of load three times.
1. The measurement results of load mass and reading on ruler
Raw Data MeasureMass of load m/(kg)? m ????? ± 0,001 kgReading on ruler by load increasing l/(m)? l=± 0,005 m10,0500,04220,1000,05630,1500,07140,2000,08550,2500,09960,3000,11470,3500,12880...
