Oscilloscope Sensitivity and Voltage Analysis

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The discussion focuses on calculating the amplitude and frequency of a sinusoidal voltage using oscilloscope measurements. The vertical sensitivity is 5 mV/cm, and the participant initially misinterprets this as 5 megavolts. After clarification, it is determined that the amplitude is found by multiplying the height of the wave in centimeters by the vertical sensitivity. For frequency, the participant identifies the period by measuring the waveform's length in centimeters and acknowledges that frequency is expressed in Hertz. The thread concludes with the participant successfully resolving their calculations with community assistance.
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Homework Statement


In the oscilloscope display above, the vertical sensitivity is 5mV/cm, and the horizontal sensitivity is 0.5μs/cm.
A) find the amplitude of the sinusoidal voltage
B) find the frequency of the sinusoidal voltagehttp://imgur.com/fnkXliz

Homework Equations


im not sure which equations to use for this question, if any are needed.

The Attempt at a Solution


I know you have to convert the units in this question I am assuming. It also says in the question that each block is 1cm by 1 cm, and the top of the amplitude is about 3.2 cm, but that isn't the correct answer.
 
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Ask yourself, what does it mean when it says 5 mV/cm? And so, what does it mean if you have a wave 3.2 cm tall on the screen?

Ask yourself, what does it mean when it says 0.5μs/cm? And how long (in cm) is one wave of the signal on the screen? And so, what does that mean?
 
i think it means (and correct me if I am wrong) that there is 5 megavolts (5000000 volts) in every centimeter on the y-axis. So if the wave is 3.2 cm tall would you multiple 5mV by 3.2 to get the amplitude?
 
reconrusty said:
i think it means (and correct me if I am wrong) that there is 5 megavolts (5000000 volts) in every centimeter on the y-axis. So if the wave is 3.2 cm tall would you multiple 5mV by 3.2 to get the amplitude?

First of all, mV is milliVolts, not MegaVolts... :-)

Next, I see the vertical amplitude as a little over 6 divisions tall. So the peak-to-peak amplitude will be a bit over 6 * 5mVpp. It's hard to tell if the question is asking for the peak or peak-to-peak "amplitude". Can you clarify that?

And on the horizontal axis, how many divisions between the positive peaks of the waveform (or alternately use the positive-going zero crossings to find the waveform's period)
 
reconrusty said:
http://imgur.com/fnkXliz
this is the graph posted in the question aswell

Better to show the picture:
fnkXliz.gif
 

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Okay so i got the first question right, it was asking for the peak not the peak to peak. So for the second part about finding the frequency, I counted roughly 3.2 blocks as the period, so the frequency is just 1/period right? I am not sure what to put in for units though, i thought frequency was measured in Hz, but when i try to put the answer in online it says the units are wrong... any help?
 
3.2 cm is how much time ?
 
Problem solved! thank you guys so much for the help :)
 
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