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Set 9 Problem number 15
Determine the restoring force constant for a simple pendulum of length 1.25
meters and mass .9 kilograms, using the knowledge that for small displacements from
equilibrium the restoring force on a simple pendulum is in the same proportion to its
weight as its displacement from equilibrium is to its length. Determine the period
of motion for this pendulum.
Choose any mass and recalculate your result. Compare with your previous
result and explain why the results should compare as they do.
The restoring force is related to the weight ( .9 kg)(9.8 m/s ^ 2) = 8.82
Newtons of the pendulum.
- For a simple pendulum for small displacements from equilibrium, restoring force
is in the same proportion to weight as displacement to length.
- The displacement is 1/10 of the length, so the restoring force is very close to
1/10 of the weight, or .882 Newtons.
For a spring, the restoring force has magnitude F = ky, where k is the spring
constant and y the displacement from equilibrium.
- By analogy, here we see that if the pendulum is displaced distance x from
equilibrium, the magnitude of the restoring force is F = kx, with k standing for the
restoring force constant.
- It follows easily that k = F/x, where F is the force at distance x from
equilibrium.
- Here we have force .882 Newtons when the displacement is 1.25/10
meters.
- Thus the force constant for this pendulum is k = .882 Newtons/(
1.25/10 meters) = 7.056 Newtons/meter.
From the force constant we can determine that the angular frequency is `omega =
`sqrt(k/m) = `sqrt[( 7.056 Newtons/meter)/( .9 kilograms)] = `angFreq radians / second.
- The period, which we desire, is the time required for one complete revolution of
2 `pi radians.
- At `angFreq radians/second, the time required for one revolution is thus 2 `pi
radians/( 0 / second) = 0 seconds.
When you do the same set of computations for a mass of your choosing, you get a
different restoring force and restoring force constant, but you get the same period.
- This occurs because both the values of k and of m are directly proportional to
the mass of the object.
- It follows that when we divide k by m we must always get the same result, giving
us the same angular frequency.
To see that k is proportional to m, note that k is obtained by dividing the
restoring force by 1/10 the length of the pendulum; the restoring force is just 1/10 of
the weight of the pendulum, which is proportional to the mass. Thus the force constant is
proportional to the mass.
When we compute `omega = `sqrt(k/m), we divide one quantity proportional to the
mass by another. This leaves the result independent of the mass.
More generally, the weight is mg, where g is the acceleration of gravity. The
restoring force at a distance of L/10 from equilibrium (where L is the length of the
pendulum) is very nearly equal to 1/10 of the weight, or mg / 10.
- The restoring force constant is thus very nearly k = (mg/10) / (L/10) = mg/L.
- Note that the 10 divides out; it wouldn't have mattered whether we used 1/10 of
the length or 1/100 or 1/37.893 of the length, the proportion would have divided out and
we would have obtained the same result for k.
We can therefore generalize and say that for a pendulum with mass m and length
L, the restoring force constant is mg / L.
- When we compute the angular frequency, we obtain `omega = `sqrt(k/m) =
`sqrt[(mg/L) / m].
- Dividing m by m leaves us with `omega = `sqrt(g/L), which does not depend on the
mass.
It turns out that the frequency of a pendulum depends only on its length.
- Recall that the original assumption was that the amplitude of the pendulum's
motion (the maximum distance from equilibrium) is small compared to the length. The
conclusions reached here apply only to this situation.
You should know the process used here.
In short brutal language, the process goes like this:
- Restoring force proportional to distance.
- Force constant mg/L.
- Angular frequency `sqrt(k/m) = `sqrt(g/L).
- Hmm! Period not depend on mass! Only on length.
- Can test. Make pendulums. Works.
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