1
$\begingroup$

These are the results I obtained:

enter image description here

The blue line is the x-axis, i.e. when I let the mass hang without it swinging. The only thing damping it is air. The force measured, is the force exerted by the bob.

enter image description here

Above is a picture of the set-up (not very good, sorry). The set-up was a mass-on-string. Mass: 1.82N Length of string (approximately): 1.7m

$\endgroup$
10
  • $\begingroup$ What's the source of the damping? Also note that the size of the force is not the only difference between forward and backwards. $\endgroup$
    – garyp
    Oct 17, 2016 at 15:26
  • 1
    $\begingroup$ Presumably your spring doesn't obey Hooke's law (specifically the symmetry requirement). $\endgroup$
    – lemon
    Oct 17, 2016 at 15:53
  • 1
    $\begingroup$ How come the force is never zero? This would mean that your system does not oscillate around the equilibrium position. Can you specify the setup? $\endgroup$
    – nasu
    Oct 17, 2016 at 17:38
  • $\begingroup$ Each cycle looks quite symmetrical to me. I think what you are observing is a small decrease in period as the amplitude of oscillations gets smaller. This is to be expected of a real pendulum or a real spring which (as lemon says) departs from Hooke's Law as amplitude increases. It would be useful to know what your setup is and what 'force' you are measuring. $\endgroup$ Oct 17, 2016 at 18:58
  • $\begingroup$ Yes, but somehow the negative amplitude of each cycle is smaller than the positive amplitude. I'm sure there's a straight forward answer to this problem but i'm just not picking up on it. Maybe involving the mass of the pendulum... ? $\endgroup$
    – Anna
    Oct 17, 2016 at 20:41

2 Answers 2

2
$\begingroup$

I do not see any asymmetry in your graph. What I do see is that as the amplitude of oscillations decreases then so does (1) the period and (2) the average tension in the string ("force") during one oscillation.

(1) Although for small amplitudes $\theta_0$ (in radians) the period $T$ is approximately constant (independent of $\theta_0$), for larger amplitudes the period is longer :
$T=2\pi\sqrt{\frac{L}{g}}[1+\frac{1}{16}\theta_0^2+\frac{11}{3072}\theta_0^4+...]$....(eqn 1)

This is because the restoring force is proportional to $\sin\theta$ rather than $\theta$. When the amplitude gets to $90^{\circ}$ the period is almost $20$% above its small-angle value.

(2) The tension $F$ in the string ("force") is the sum of the component of the weight $mg$ of the bob along the string plus the centripetal force required to keep the bob moving in a circle of radius $L$ with speed $v$ :
$F=mg\cos\theta+\frac{mv^2}{L}$....(eqn 2)

Since angular displacement $\theta=\theta_0\sin(\omega t)$, force $F$ is not a strictly sinusoidal function. Here $\omega=\frac{2\pi}{T} \approx \sqrt{\frac{g}{L}}$ is the angular frequency of the pendulum. $v$ is related to displacement $\theta$ by
$v=L\frac{d\theta}{dt}$
$\frac{v^2}{L}=L\omega^2\theta_0^2\cos^2(\omega t)=g(\theta_0^2-\theta^2)$
therefore
$F=mg(\cos\theta+\theta_0^2-\theta^2)$...(eqn 3).

Note that $\omega$ is not equal to $\frac{d\theta}{dt}$. $\omega$ is related to the period of the pendulum and is constant if amplitude is constant, whereas $\theta$ is angular displacement and $\frac{d\theta}{dt}$ is angular velocity of the bob, both of which vary sinusoidally during one oscillation.

Tension is greatest in each cycle when the bob swings through the vertical position $(\theta=0)$ :
$F_{max}=mg(1+\theta_0^2)$....(eqn 4)
Tension is least in each cycle when the bob reaches the maximum displacement $(\theta=\theta_0, v=0)$ :
$F_{min}=mg\cos\theta_0$....(eqn 5)

As amplitude decreases, $F_{max}$ decreases but $F_{min}$ increases. When the pendulum stops $(\theta_0=0,v=0)$ the tension is $F=F_{max}=F_{min}=mg=1.82N$, which is the value for the blue line on your graph.

enter image description here

The graph above shows how the force in Newtons varies with time in seconds, starting from an amplitude of 45 degrees, with a value for damping of $\gamma=0.05$ (which is considerably higher than in your results). I have used the small-angle value for the period throughout. Note that the average force decays from about 2.1N to about 1.82N.


You have not explained the purpose of your experiment, so it is not clear how you intend to make use of your results, which show how $F$ varies with time $t$ rather than with amplitude $\theta_0$.

(i) You could investigate the damping by rearranging eqns 3 & 4 to find how $\theta_0(t)$ varies with time. The 2 eqns should give the same values for $\theta_0$; you could calculate both and use the average.

If the damping force is proportional to the speed of the bob (which is what is expected for air resistance at 'low' speeds) you should get an exponential decay $\theta_0 \propto e^{-\gamma t}$ where $\gamma$ is the decay constant. So a plot of $\ln(\theta_0)$ against $t$ should be a straight line with -ve slope.

(ii) You could investigate how the period $T$ varies with amplitude $\theta_0$, by comparing your results with the prediction in eqn 1. Again you need to obtain $\theta_0(t)$ from the values of $F_{max}$ and $F_{min}$.

It is probably best to average the period over a few oscillations. Note also that force $F$ reaches a maximum twice in each oscillation of the pendulum, so the period of the pendulum will be twice that in your graphs.

$\endgroup$
16
  • $\begingroup$ So that means that the blue line is not actually my x-axis for the other graphs. $\endgroup$
    – Anna
    Oct 17, 2016 at 22:06
  • $\begingroup$ @Anna : Sorry I don't know what other graphs you need to plot. $\endgroup$ Oct 17, 2016 at 22:26
  • $\begingroup$ I mean the other graphs on the graph. The green, red, pink and orange line ... $\endgroup$
    – Anna
    Oct 18, 2016 at 11:07
  • $\begingroup$ Those are the same graph of the pendulum, but at later times. The average value of force during one oscillation (what you are calling the x axis) decreases as amplitude decreases. The oscillations should not be symmetrical about the blue line, as you assumed. There is nothing going wrong with your graph, no technical problem caused by the equipment or the apparatus. The graph is exactly what you should expect : as amplitude $\theta_0$ decreases then $F_{max}$ decreases, $F_{min}$ increases, and average force $\frac12(F_{max}+F_{min})$ decreases. $\endgroup$ Oct 18, 2016 at 15:04
  • $\begingroup$ What formula did you use for the period? The link doesn't work. $\endgroup$
    – Anna
    Oct 18, 2016 at 17:19
-1
$\begingroup$

Because your damping arrangement produces an inconsistent damping force between one direction and the other.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.