Video Transcript
A pendulum has a length of 0.500
meters. What is the period of the
pendulum? Use a value of 9.81 meters per
second squared for the local acceleration due to gravity. Give your answer to three
significant figures.
So in this question, we’re
considering a pendulum, and we’re being asked to find its period. For any oscillating system, the
period refers to the amount of time taken to complete a full cycle. For a pendulum, that means the time
taken for it to swing from its starting position to the other side and then back to
its starting position. In this case, we’re told that our
pendulum has a length of 0.500 meters. And we’re also told that the local
acceleration due to gravity is 9.81 meters per second squared, which is the standard
gravitational acceleration for an object near the surface of the Earth.
Now, it may not seem like we’ve
been given much information about this pendulum. For example, we have no idea what
its mass is. However, this is all the
information we need to calculate its period. In fact, we can recall that the
period of a pendulum is given by this formula. 𝑇 is approximately equal to two 𝜋
times the square root of 𝐿 over 𝑔, where 𝑇 is the period of the pendulum, 𝐿 is
the length of the pendulum, and 𝑔 is the gravitational acceleration experienced by
the pendulum.
Note that in this formula, 𝑇 is
only approximately equal to the expression on the right-hand side. This is because the formula is
derived using a number of approximations. Importantly, it uses a small-angle
approximation, which means it’s only accurate if the initial angle of the pendulum
from the vertical, 𝜃 zero, is much smaller than one radian. The bigger the initial angle is,
the more 𝑇 will vary from the expression given on the right. So in our answer to this question,
we’re assuming that the pendulum only swings over a small angle.
In this question, we’re told that
the length of the pendulum is 0.500 meters and the local acceleration due to gravity
is 9.81 meters per second squared. So all we need to do to calculate
the period of the pendulum is substitute these values into this formula. Substituting them in gives us two
𝜋 times the square root of 0.500 meters divided by 9.81 meters per second
squared.
At this point, it’s worth
mentioning that this formula only works when we use the magnitude of the
acceleration. It’s fairly common in physics
problems to define upward acceleration as positive and downward acceleration as
negative. But when we’re using this formula,
it’s important that we don’t use a negative value for acceleration.
If we type all of this into our
calculator, we obtain a value of 1.418 and so on, which, rounded to three
significant figures, is 1.42. Since the period is a measure of
time, we would expect our answer to have units of time. And if we look at the units we’ve
used in our calculation, we can see that this is the case. We’re dividing the length, which is
a quantity in meters, by an acceleration, measured in meters per second squared. And we’re then taking the square
root of this quantity. We can see that in this fraction,
the factor of meters on the top and the bottom will cancel, leaving us with the
square root of one over one over second squared.
Now, when we have one over one over
a quantity, in other words, the reciprocal of the reciprocal of a quantity, this
actually just leaves us with whatever quantity it was that we started with, in this
case second squared. And the square root of second
squared is of course seconds, which shows us that our answer is 1.42 seconds. And this is the answer to the
question. If a pendulum has a length of 0.500
meters and the local acceleration due to gravity is 9.81 meters per second squared,
then the period of the pendulum is 1.42 seconds.
