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 we’ve been 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 units of volts, plotted on the 𝑦-axis and the time,
in units of 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 output potential difference, we first need to find the period.
The period is simply the time taken
for one complete cycle of the oscillation to occur. We can use the graph to work this
out. We could look at any oscillation
cycle shown on this graph, but it’s probably easiest to look at this first one
here. 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 cycle in our working. Otherwise, we wouldn’t calculate
the right value for the period.
The period, or 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 read
these values from the graph. The scale marked on the horizontal
axis tells us that 10 of these vertical grid lines corresponds to 100
milliseconds. So we know that each individual
line must represent 10 milliseconds.
The oscillation begins at a time of
zero milliseconds, and it ends here at the second grid line. Since one vertical grid line
corresponds to 10 milliseconds, then the end of this first cycle after two grid
lines must correspond to 20 milliseconds. The period of oscillation is equal
to the time difference between these two times, and so the period is equal to 20
milliseconds minus zero milliseconds. This is simply equal to 20
milliseconds.
However, before we can use this
value to calculate the frequency of the oscillation, we first have to convert it
into units of seconds. Recall that the unit prefix milli-
is equivalent to a factor of 10 to the negative three. So 20 milliseconds is equal to 20
times 10 to the negative three seconds, which we can write more simply as 𝑇 equals
0.02 seconds. So the oscillation period of the
output potential difference is 0.02 seconds.
Let’s clear some space on screen so
we can substitute this into our equation for frequency. Substituting our value for the
period into this equation, we have that the frequency of the oscillation is equal to
one divided by 0.02 seconds.
Before we calculate this number,
let’s take a moment to think about the units of the quantities. We have no units in the numerator
of this expression, and in the denominator we have units of seconds. Overall, this gives us units of
inverse seconds or seconds to the negative one.
However, frequency is more 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 by units of hertz. Now, all that’s left to do is to
calculate the value of one over 0.02 hertz. This gives us our final answer of
50 hertz. We have found then that the
frequency of the output potential difference of the generator is equal to 50
hertz.
