### Video Transcript

The graph below shows the output
potential difference of a generator over time. What is the frequency of the output
potential difference?

To answer this question, we need to
use the graph given to work out the frequency of the output potential difference of
a generator. We can see that this graph has the
output potential difference in volts plotted on the 𝑦-axis and the time in
milliseconds plotted on the 𝑥-axis. Recall that the frequency of an
oscillation can be calculated using the formula frequency equals one divided by the
period of oscillation. This is often written as 𝑓 equals
one over capital 𝑇, where 𝑓 is the frequency and capital 𝑇 is the period.

So, before we can find the
frequency of the potential difference, we first need to find the period. The period is simply the time taken
for one complete oscillation to occur. We can use the graph to work this
out. We could look at any oscillation
cycle shown on the graph, but it’s probably easiest to look at this first one. Notice that we’re including this
starting point here, this maximum value, this minimum value, and this zero point
here in one single oscillation. It’s important that we include the
whole oscillation cycle in our working; otherwise, we wouldn’t calculate the right
value for the period.

The time taken for this oscillation
to occur is simply equal to the time difference between this point here on the time
axis and this point here. So, all we really need to do is to
read these values from the graph. The scale marked on the horizontal
axis tells us that five of these vertical grid lines corresponds to 500
milliseconds. So we know that each vertical line
must represent 100 milliseconds. This oscillation begins at a time
of zero milliseconds and ends at the second vertical grid line, which corresponds to
a time of 200 milliseconds. The period of oscillation is equal
to the time difference between these two times. That is, the period capital 𝑇 is
equal to 200 milliseconds minus zero milliseconds. This is simply equal to 200
milliseconds.

We have now found the period of the
oscillation. But before we can use this value in
this equation to calculate the frequency of the oscillation, we first have to
convert it into units of seconds.

Recall that the unit prefix
lowercase m, or milli-, is equivalent to a factor of 10 to the power of negative
three. So, 200 milliseconds is equal to
200 multiplied by 10 to the power of negative three seconds, which we can write more
simply as 0.2 seconds. So, the period of the output
potential difference is 0.2 seconds. Let’s now clear some space on
screen so we can substitute this value into our equation for the frequency.

Substituting our value for the
period, we have that the frequency of this oscillation is equal to one divided by
0.2 seconds. Before we calculate this number,
let’s take a moment to think about our units. We have no units in the numerator
of this expression. And we have units of seconds in the
denominator. Overall, this gives us units of
inverse seconds, or seconds to the negative one. However, frequencies are most
commonly expressed in units of hertz. One hertz is equal to one inverse
second. So, we can just replace the units
of inverse seconds in our expression with units of hertz. The frequency 𝑓 is equal to one
over 0.2 in units of hertz.

Now, all that’s left is to
calculate the value of one over 0.2 hertz. This gives us our final answer of
five hertz. We have found, then, that the
frequency of the output potential difference is equal to five hertz.