Simple Harmonic Motion - Period of a Pendulum

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  • #1
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We are talking about SHO in physics and the forumula for period T of a pendulum was introduced - T=2*pi*(L/g)^.5 where L is the length of the pendulum arm/string and g is acceleration due to gravity. the instructor also said this was an approximation that was only accurate for small angles. so i thought what about large angles? - like if it is dropped from a horizontal position.

I did some work and derived my own general formula for any angle. i dont know how correct it is but here is what i came up with:

T = 2*Theta*[ (2*L)/(g*(1-cos(Theta))) ] ^ (1/2)

Theta is the starting angle that it is dropped from - displaced from the vertical - it is zero at the equilibrium position (just hanging straight down), and it is pi/2 or 90 deg if you hold it out horizontally.

so to test it mathematically against the other formula i made an excel table with values of theta from 0 to pi/2 in increments of 0.1 rad . in one column i had the value from my formula and the next column from the books formula. they were different but i noticed some connections. for example when g=9.8 and L=1 and theta = pi/2, my formula came up with 1.42s and the other came up with 2s. however the books formula is only an approximation for small angles so this is expected. yet when theta = 0.1 there is still discrepency - mine=1.28s , book=2s. I expected that as theta got smaller my formula and the books would have closer and closer numbers, but i was wrong.

then i made a column for percent difference between the two. it ranged from 36 (theta=0.1) to 30 percent (theta=pi/2). however i noticed something interesting. no matter what i changed the length L to, the percent difference column remained constant.

:bugeye: any thoughts?
 
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  • #2
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Conditions for SHM:

Fixed amplitude

Elasticity and Inertia

Force should always direct towards the centre of oscillation

Fixed Time Period

"Must be Straight line"

WE TAKE SMALL ANGLES WHILE DOING SHM PROBLEMS LIKE THE CASE OF PENDULUM
.THE REASON FOR THAT IS: FOR A GOOD APPROXIMATION SMALL ANGLES MEAN STRAIGHT LINE MOTION OF THE BOB ATTACHED TO THE THREAD . When we take to larger angles the motion is not SHM. It is called "osscilation" and not "SHM" . Oscillation means to and fro motion which may or may not be periodic. SHM is a special case of oscillation which undergoes in some conditions as mentioned above. Moreover larger angles means damping effect due to air and hence no SHM.
 
  • #3
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Dr. Brain,

Your condition that SHM "Must be Straight line" seems overly restrictive. The reason that SHM for a simple pendulum is limited to small angles is that the approximation of a linear restoring force is only accurate for small angles.

In fact, it's possible to construct a pendulum which exhibits SHM even for large angles, where its motion is nowhere near along a straight line (but it's along a cycloid, not a circle).
 
  • #4
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im disregarding all friction or any other dampening force. even with a larger angle it would still oscillate with a constant period and amplitude etc. so it is still SHM, it just can be described by the common pendulum formula because the error would be to great. Since even at small angles, my formula produces different numbers than the other, im trying to figure out where the discrepency is coming from. maybe if i showed my work...

i started with the basic kinematics equation converted to rotational motion:
Theta=(1/2)Alpha*t^2, where Theta is angular displacement, Alpha is angular acceleration, and t is time. I solved for t to get:
t = (2*Theta/Alpha)^(1/2)
Now if i calculate the time it takes a simple pendulum to fall to equilibrium from some starting angle, if i double that I will have the period of oscillation.
However, the angular acceleration is not constant. If i find the average acceleration between the starting angle and equilibrium i can plug that back into the formula.
I have the formula:
Alpha=Torque/I where I is moment of inertia. Since moment of inertia is constant, torque is what is changing.
The torque at some instant is Torque=mgLsin(Theta) , where L is the length of the pendulum arm and Theta is the angle from the vertical it is at at that instant.
To find the average torque, I integrated mgLsin(Theta) with respect to Theta, from 0 to Theta. Then I divided by Theta to get the average torque.
I found average torque to be:
Torque=(mgL(1-cos(Theta)))/Theta
After i substituted to find average angular accel and simplified, i got:
Alpha=g(1-cos(Theta))/Theta*L

Then i plugged that back into the original formula for time t, multiplied by 2, and simplified to get period:

T = 2*Theta*[ (2*L) / (g*(1-cos(Theta))) ] ^(1/2)
 
  • #5
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Use the series expansion for Sin(x) (power law series) instead of simply Sin(x)

That'll work for large angles.

[tex] \sin\theta = \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} ....[/tex]


Use this in the differential equation
[tex]m\frac{d^2 \theta}{dt^2} = -\frac{g}{L}\sin\theta[/tex]

Its a fun differential equation to solve... I'd suggest doing it numerically using a computer

When you do this you'll see that the solutions will contain harmonics of [tex] \omega_0 [/tex]
 
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  • #6
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ok but it seems to me that my formula should still work even tho i did it a different way...
 
  • #7
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I dont think your formula will accurately describe the motion of a pendulum for large angle oscillations
 
  • #8
dextercioby
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There it is.

[tex] T=2\pi\sqrt{\frac{l}{g}} \ _{2}F_{1} \left(\frac{1}{2},\frac{1}{2},1;\kappa^{2}\right) [/tex]

,where [itex] \kappa^{2}=\sin^{2}\frac{\alpha}{2} [/itex]

,where [itex] \alpha [/itex] is the angular amplitude.


It is obtainable from the law of conservation of mechanical energy.


Daniel.
 
  • #9
dextercioby
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If u prefer,u can use the Legendre's complete elliptical integral of first kind [itex] K(x) [/itex] and then the result is trivial

[tex] T=4\sqrt{\frac{l}{g}} K\left(\kappa^{2}\right) [/tex]


Daniel.
 
  • #10
dextercioby
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Comparing the 2 formulas i've given u for the period of oscillation,u can retrieve the relation #3 from here


Daniel.
 
  • #11
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what is flawed with my calculation of the average torque? I just have basic physics knowledge, and my highest math is calc2, so your eliptical integral went over my head, alot of the stuff in your first formula is unfamiliar to me - starting at the first subscript 2, im not sure what the rest means.
 
  • #12
dextercioby
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The torque is time dependent (the angular acceleration is time dependent).I dunno what average u computed there,it surely wasn't the right one.

Daniel.
 
  • #13
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the torque is really only dependent on the angle the pendulum is at (max when horizontal and zero when vertical). it does change with time but the change isnt caused by the time change. i computed the average by integrating torque as a function of theta from zero to the starting angle, and then divided the result by the starting angle to find the average torque. this is just the concept of finding the average of a function. but i did find a calculation error, my formula is only for half the period so the 2 should actually be a 4 to have the full period.
 
  • #14
dextercioby
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You're in a complete missuderstanding.

[tex] \vec{\tau}(t)=:I\vec{\alpha}(t) [/tex]

and the torque's dependence of time is explicit.

The angular acceleration can be computed solving the ODE

[tex]\frac{d^{2}\vartheta\left(t\right)}{dt^{2}}+\frac{g}{l}\sin\vartheta\left(t\right)=0 [/tex]

Daniel.
 
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  • #15
dextercioby
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That ODE is subject to the initial conditions

[tex] \vartheta (0)=\vartheta_{0} [/tex]

[tex] \omega (0)=:\dot\vartheta (0) =0 [/tex]

And its solutions are found by reversing the formula given in the post #9 using Jacobi's elliptic functions,[itex] \mbox{sn}(x) [/itex],if i'm not mistaking.

Daniel.
 
  • #16
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For Large amplitude oscillations I get the following formula:

[tex] T = 2\pi\sqrt{\frac{L}{g}}(1 + \frac{1}{16}\theta^2 + \frac{11}{3072}\theta^4 ...)[/tex]

theres some first 3 terms to the series way of doing it. I didnt work out any others, but I'm pretty sure this is correct. (if you just want the formula)
 
  • #17
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hmmm....i was looking through the books proof of the regular formula and where they assumed that sin (theta) = theta, i left it in and ended up with a pretty simple formula that should work for all angles:

T=2*pi*(L*sin(theta)/g*theta)^(1/2)

i put this into my excel table and compared it with the original formula from the book and it makes alot of sense.
 
  • #18
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Felix83,

"i left it in and ended up with a pretty simple formula that should work for all angles"

If gregmead's formula is correct (post just before yours), then yours can't be because his T is an increasing function of theta; yours is a decreasing function of theta.

But more to the point, why do you think your formula should work for all angles?
 
  • #19
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I think your error is coming from taking the average of something, I'm not sure how you came up with this formula...
 
  • #20
HallsofIvy
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Felix83 said:
hmmm....i was looking through the books proof of the regular formula and where they assumed that sin (theta) = theta, i left it in and ended up with a pretty simple formula that should work for all angles:

T=2*pi*(L*sin(theta)/g*theta)^(1/2)

i put this into my excel table and compared it with the original formula from the book and it makes alot of sense.
You left sin(θ) in? Then how did you get that period? Do you understand that then the equations have solutions that are not even periodic?
 
  • #21
Integral
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I have a VB program all set up to calculate the motion of a non linear pendulum using a Runga Kutta method.

for a initial displacement of a 1 meter pendulum of 1.2 radians I get T= 2.2 secs.

using Gregmeads formula I get, for that starting displacement, T= 2.1 sec

Using felix83's formula T=1.76s

Greg you may want to generate another term or 2.

Felix83..... back to the drawing board! :smile:
 
  • #22
krab
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If we define
[tex]
\lambda = {1 - \sqrt{\cos (\phi_{\rm max}/2)}\over 1 +
\sqrt{\cos (\phi_{\rm max}/2)}}
[/tex]
then to a very good approximation
[tex]
T/T_0 =(1+\lambda)^2
[/tex]
This estimate is good to within 10% up to [itex]\phi_{\rm max} =
178^\circ[/itex] and to within 1% up to [itex]\phi_{\rm max} = 168^\circ[/itex].
And indeed for [itex]\phi_{\rm max} = 1.2[/itex] radians, T/T_0=1.098. Since [itex]T_0=2\pi\sqrt{1/9.8}=2.007[/itex] second, T = 2.2 sec as Integral finds.

This approx can be found in E.T. Whittaker and G.N. Watson Modern Analysis Sec. 21.8 (1927).
 
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  • #23
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i see where i went wrong, whoever said torque is a fucntion of time, is right. well it depends on angle as well as time, but i assumed it depended on angle alone. however, since it is already moving faster towards the bottom of its fall, it experiences this torque over a shorter time interval so it would change the average. so my average torque calculation is what is wrong. the second formula i posted where i followed the books proof, except i left sin theta in, is the right formula. In my excel spread sheet, for g=9.8, L=1, and theta=0.1 rad, the books original formula got 2.007s and my altered formula got 2.005s. the numbers are very close for small angles and differ more as the angle gets bigger, which is exactly what is expected. for theta = pi/2 i got T=1.6s.
 
  • #24
Integral
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Felix you need to read what has been posted a bit closer. You may note that the correct formulation has the period increasing with initial amplitude. Yours is decreasing, this is not correct.

Did you compare your result to the formula posted by Krab? Perhaps you should use that as a benchmark. When you find agreement with that formula you will be onto something.

BTW for an initial displacement of 1.57 rad T=2.36s
 

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