An alternator (AC generator) is rotated at a constant frequency fff, and its electrical output is monitored on an oscilloscope as shown in the trace below.
The settings of the oscilloscope are kept constant. If the frequency of rotation of the alternator is doubled to 2f2f2f, what will be the new amplitude (peak voltage) and the new period of the displayed wave, measured in grid divisions?
Amplitude=2 divisions,Period=2 divisions
\text{Amplitude} = 2\text{ divisions}, \quad \text{Period} = 2\text{ divisions}
Amplitude=2 divisions,Period=2 divisions
Amplitude=4 divisions,Period=8 divisions
\text{Amplitude} = 4\text{ divisions}, \quad \text{Period} = 8\text{ divisions}
Amplitude=4 divisions,Period=8 divisions
Amplitude=4 divisions,Period=2 divisions
\text{Amplitude} = 4\text{ divisions}, \quad \text{Period} = 2\text{ divisions}
Amplitude=4 divisions,Period=2 divisions
Amplitude=8 divisions,Period=2 divisions
\text{Amplitude} = 8\text{ divisions}, \quad \text{Period} = 2\text{ divisions}
Amplitude=8 divisions,Period=2 divisions