Problem: The restoring force for a pendulum, when the displacement of the pendulum from its equilibrium position is small compared to its length, is in the same proportion to its weight as its displacement from equilibrium is to its length. What therefore is the restoring force on a pendulum of length 2 meters and mass 7 kilograms when the mass is displaced 1/10 of its length from its equilibrium position? What is the restoring force constant, analogous to the spring constant of a spring, for this pendulum? What should be the period of its motion? Recompute for a pendulum of the same length, but with a mass of your choosing. What do you obtain for the period? Comment in your summary on why the result comes out as it does.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: The restoring force is related to the weight ( 7 kg)(9.8 m/s ^ 2) = 68.6 Newtons of the pendulum. The displacement is 1/10 of the length, so the restoring force is 1/10 of the weight, or 6.86 Newtons. `p`pFor 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. `p`pIt follows easily that k = F/x, where F is the force at distance x from equilibrium. Here we have force 6.86 Newtons when the displacement is 2/10 meters. Thus the force constant for this pendulum is k = 6.86 Newtons/( 2/10 meters) = 34.3 Newtons/meter. `p`pFrom the force constant we can determine that the angular frequency is `omega = `sqrt(k/m) = `sqrt[( 34.3 Newtons/meter)/( 7 kilograms)] = 2.213 radians / second. `p`pThe period, which we desire, is the time required for one complete revolution of 2 `pi radians. At 2.213 radians/second, the time required for one revolution is thus 2 `pi radians/( 2.213radians / second) = 2.839 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. m is of course proportional to itself.
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 1/10 of the weight, or mg/10.
The restoring force constant is thus 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.