Acceleration due to gravity using a simple pendulum.?

Question

Labatory conditions; Apparatus, A peice of cord about 2m in length, a fixed pivot point,a symetrical pendulum bob, a measure tape and stop watch. measure the cord ar 5 different lengths and swing pendulum through a small arc of 10 degrees for 10 oscillations, repeat twice for average time and plot findings. CONLUSION this is the bit i need help with! outline the reasons why this method only gives an approximate value, no matter how ACCURATELY the experiment is performed.

Answer 1

If you use a heavy enough weight then the effects of air are minimal.

Answer 2

This is because the effect of gravity isn't constant. When the pendulum is at the bottom of the swing gravity actually has no effect on the pendulum, it is centrifugal force (or is it centripetal ... whatever) which is keeping the string tight. Thus one can say the only time gravity has 100% effect is when the pendulum has swung to either extreme and has stopped. At that moment the effect of gravity is 100%.
Addition:
When the pendulum is at 90 deg to the vertical then you have (Sine 90 deg =) 100% pull on the weight, when it is a 45 deg gravity has (Sine 45 deg = ) about 70% pull. Your experiment said 10 degrees, which means gravity has only 17% influence on the weight. The other 83% goes into the string.

Answer 3

There is not only air resistance friction but also there is friction within the measurement devices. Internal friction in the cord, as well as friction in the timing devices, and calibration error of the measurement devices. Additionally, the measurements will be affected by the position of the moon, as well as the orientation and direction of the apparatus in relation to the earths magnetic field if magnetic or ferrous materials are used in either the swing apparatus or the measurement apparatus. There also may be interference from the rotation of the earth via centrifugal force and momentum of the pendulum in relation to these forces. Again the orientation of the apparatus would effect the measurement. I hope this helps!

Answer 4

Try using a weightier and if it's possible smaller, so choose the highest density possible(Pd is better than Fe for example), and try to choose a sphere, those things may help you to reduce air drag respect to inertia, so you reduce approximation.
But except this and other little effects, in experiments you usually don't get very accurate values, because the constant found by experimental way (g or the constant between cal and joule) has been obtained making a lot (not 2 but 100 or 1000) of measures and choosing the mean value, sometimes with statistic procedures.

Answer 5

Your cord is too short and the bob needs to be heavy to virtually eliminate air resistance. They have a pendulum in the Science Museum in London and it's (can't remember exactly) about 30 meters long. And 10 oscillations is not enough to compensate for errors in your timings. Try 10,000. But note that the pendulum will swing round as the earth rotates.

Answer 6

ok This assumes you know about simple harmonic motion.
Restoring force on pendulum mass= mgsin(theta)
where g= 9.81=grav field intensity,m=bob mass,theta=angle with vertical
NewtonII> F=ma (a=acceln of bob mass)
So mgsin(theta)=ma
> gsin(theta) =a
Now, this is the bit! for small angles, theta in radians = sin(theta) approx
(The error when theta is 10deg= 0.5% -check it on your calculator!)
> g(theta) = a (Which is the eqation for simple harmonic motion)
In SHM if x=bcoswt
Differentiating twice and substituting x=bcoswt
> a=d2x/dt2= -w^2x where b=amplitude, w= angular freq
Also theta= x/L approx, where x=arc length, L=pendulum length (diagram needed!)
> gx/L= w^2x in magnitude
cancel x's, subst T=2*pi/w
> T= 2*pi* sqrt(L/g)
So,(!) If you plot a graph of L vs T^2 you get a straight line gradient= g/(4*pi^2)
But, rememember that teeny approximation?
The 'real' value of sin(theta) is given by a power series involving even powers of theta, alternate (+/-) values, divided by even factorials of these powers (McLaurin exp)
So no matter how accurately you measure T, L there will always be that approximation error.
If you measured 45deg swings for example, the error in g would be about 11%!
Or, to put it succintly, the pendulum does not describe true SHM, where the restoring force is directly proportional to its displacement from its origin of motion and oppositely directed!

Answer 7

No, the period is T = 2pi SQRT { L/g} where L = length of the pendulum. At small angles, there is no dependence on the mass of the pendulum bob. g = acceleration of gravity so if you measure the period and the length you can determine g.

Answer 8

This will give u approx. result becoz the motion of the pendulam will be affected(damped) by the air of the atmosphere.. which u may not take in account. If u do the experiment in vacuum then the result will be accurate.

Answer 9

did u take into account the air's resistance.