
This problem set deals with alternating
current and frequency. 


1. 
An electric generator that produces 600
cycles in 10 sec has a frequency of 60 Hz. 


2. 
The common unit of frequency for one
cycle/s is 1 Hz. 


3. 
If a wave has a frequency of 10 cycles/s
(10 Hz), what is its period?
T = 1/f = 1/(10
cycles/sec) = 0.1 sec/cycle



4. 
Frequency is a measure of how often
events repeat themselves. The graph below is a sine wave.
If point A is the starting point, at what later points
does the wave repeat itself? That is, when does the cycle
start over again?
The cycle starts over at points E and H.



5. 
In the graph above, how many complete
cycles are shown for the wave?
Two cycles (A to E and E to H)



6. 
Given the oscilloscope below with the
indicated settings, find the
 time/division and volts/division
 period of one cycle of this waveform
 peaktopeak voltage of the signal.
 time/div is 2 ms/div and volts/div is 10
volts/div
 T = (8
div)(2
ms/div) = 16
ms
 V_{pp} = (7
div)(10
volts/div) = 70
volts



7. 
Name the type of waveform shown below.
Find its period, frequency and peaktopeak voltage.
 square wave
 T = (5
div)(0.2
sec/div) = 1
sec
 f = 1/T = 1/(1sec/cycle) = 1
Hz
 V_{pp} = (5
div)(0.1
V/div) = 0.5
V



8. 
Find the peaktopeak voltage, period
and frequency of the sine waveform.
 V_{pp} = (6
div)(2
V/div) = 12
V
 T = (8
div)(50
msec/div)(1
sec/10^{6} msec) = 0.0004
sec (400 msec)
 f = 1/T = 1/(0.0004
sec/cycle) = 2500 Hz



9. 
Find the peaktopeak voltage, period
and frequency of the sawtooth waveform shown below. Note
that a "×10" oscilloscope probe was used to
obtain the signal. This means that the voltage must be
multiplied by a factor of ten to get the proper answer.
 V_{pp} = (8
div)(5
V/div)(×10) = 400
V
 T = (8
div)(20
msec/div) = 160
msec (0.16 s)
 f = 1/T = 1/(0.16
sec/cycle) = 6.25 Hz



10. 
Find the peaktopeak voltage and
frequency of the sine wave shown below. Note that a
"x10" oscilloscope probe was used to obtain the
signal. This means that the voltage must be multiplied by
a factor of ten to get the proper answer.
 V_{pp} = (8
div)(2
V/div)(×10) = 160
V
 f = 1/T= 1/[(10
div)(0.002
sec/div)] = 50
Hz



11. 
The electron gun in a TV picture tube
scans the width of the screen. To form an image, it
produces 15.75 ×10^{4} lines on the screen
during a 5 minute period. Find the frequency of the
electron gun.


Knowns 
Unknowns 
Equations/Remarks 
n = 157,500
lines
t = 5 min = 300 sec 
f = ? 
f = n/t 
f = n/t = (157500
lines)/(300 sec) f = 525 lines/sec

Divide the number of cycles
(lines in this case) by the amount of time it
takes. 


12. 
The oscilloscope of an engine analyzer
shows that the waveform for an electronic ignition device
on a 6cylinder engine has a frequency of 6000 pulses/min
at an engine speed of 2000 rpm. Find the
 number of times the device pulses during a 5
minute test
 period in seconds between successive pulses.


Knowns 
Unknowns 
Equations/Remarks 
f = 6000 pulses/min
t = 5 min
1 min = 60 sec 
a. n = ?
b. T = ? 
f = n/t
T = 1/f 
a. f = n/t
n = ft = (6000
pulses/min)(5
min)
n = 30,000
pulses b.
T = 1/f = 1/[(6000 pulses/min)(1 min/60
sec)]
T = 0.01 sec

Rearrange the frequency equation
to solve for the number of cycles (pulses).
Substitute in the known values and calculate the
answer. Convert the minutes to seconds to find
the period



13. 
A train of identical flat cars travels
at constant speed past an observer. The observer sees a
new flat car go by every 4 sec. What is the frequency at
which the cars pass the observer?
f = 1/T = 1/(4
sec/cycle) = 0.25 Hz



14. 
What is the period of the following
waves:
 50 Hz
 50 kHz (50,000 Hz)
 50 MHz (50,000,000 Hz)
 T = 1/f = 1/(50
cycles/sec) = 0.02 sec
 T = 1/f = 1/(50,000
cycles/sec) = 2 ×10^{5}
sec (20 msec)
 T = 1/f = 1/(50,000,000
cycles/sec) = 2 ×10^{8}
sec


