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Force analysis of a pendulum
How to change parameters?Set the initial position
Click and drag the left mouse button
The horizontal position of the pendulum will follow the mouse Animation starts when you release the mouse button
- Adjust the length
dragging the pointer (while > holding down the left button)
from the support-point (red dot) to a position that sets the length you want.
Animation starts when you release the mouse button
- Change gravity g
Click near the tip of the red arrow,
and drag the mouse button to change it (up-down).
- Change the mass of the bob
Click near the buttom of the black stick,
and drag the mouse button to change it (up-down).
Information displayed:
1. red dots: kinetic energy K = m v*v /2 of the bob 2. blue dots: potential energy U = m g hof the bob
Try ro find out the relation between kinetic energy and pontential energy! 3.black dots (pair) represent the peroid T of the pendulum
move the mouse to the dot :
will display information for that dot in the textfield
Click
show checkbox to show more information
blue arrow(1): gravity green arrows(2): components of gravity red arrow
(1): velocity of the bob
Try to compare velocity and the tangential component of the gravitional force!
The calculation is in real time (use Runge-Kutta 4th order method). The period(T) is calculated when the velocity change direction.
You can produce a period verses angle ( T - X ) curve on the screen,just started at different positions and wait for a few second.
Therotically, the period of a pendulum $T=sqrt{g/L}$.
Purpose for this applet:
1. The period of the pendulum mostly depends on the length of the pendulum and the gravity (which is normally a constant)
2. The period of the pendulum is independent of the mass.
3. The variation of the pendulum due to initial angle is very small.
The equation of motion for a pendulum is $ frac{d^2 heta}{dt^2}=-frac{g}{L}, sin heta$
when the angle is small $ heta << 1$ ,$sin hetaapprox heta$
so the above equation become $frac{d^2 heta}{dt^2}approx-frac{g}{L}, heta$
which imply it is approximately a simple harmonic motion with period $T=2pi sqrt{frac{L}{g}}$
What is the error introduced in the above approximation?
From Tayler's expansion $sin heta= heta-frac{ heta^3}{3!}+frac{ heta^5}{5!}-frac{ heta^7}{7!}+frac{ heta^9}{9!}-frac{ heta^11}{11!}+...$
To get first order approximation, the error is $frac{ heta^3}{3!}=frac{ heta^3}{6}$
So the relative error (error in percentage)= $frac{ heta^3/6}{ heta}=frac{ heta^2}{6}$
If the angle is 5 degree, which mean $ heta=5*pi/180approx=5/60=1/12$
So the relative error is $ frac{ heta^2}{6}=1/(12^2*6)=1/(144*6)=1/864approx 0.00116$
For angle=5 degree , the relative error is less than $0.116%$
For angle=10 degree , the relative error is less than $0.463%$
For angle=20 degree , the relative error is less than $1.85%$
So the period of the pendulum is almost independent of the initial angle (the error is relatively small unless the angle is much larger than 20 degree- for more than 2% error).