Question Video: Determining the Frequency of the Output Potential Difference of a Generator | Nagwa Question Video: Determining the Frequency of the Output Potential Difference of a Generator | Nagwa

Question Video: Determining the Frequency of the Output Potential Difference of a Generator Physics • Third Year of Secondary School

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

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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.

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