Lissajous figure and oscilloscopes problem

  • Thread starter Thread starter figs
  • Start date Start date
  • Tags Tags
    Figure
AI Thread Summary
Lissajous figures on oscilloscopes can transition from straight lines to circles based on the phase relationship between two input signals; in-phase signals create straight lines, while signals that are 90 degrees out of phase produce circles. The horizontal sweep rate is influenced by the time/div control setting, which determines how many cycles of a periodic signal are displayed on the screen. A sawtooth signal is typically used for horizontal deflection, while applying a sinusoidal signal alters the sweep rate, leading to various Lissajous patterns. The frequency ratio of the horizontal to vertical signals also affects the resulting figure, with higher ratios yielding more complex shapes. Understanding these principles is essential for effectively using oscilloscopes and interpreting Lissajous figures.
figs
Messages
27
Reaction score
0
i'm learning about oscilloscopes, and i don't understand
why this Lissajous figure can be go from a straight line to
a circle. Also, this sweep time/div control setting, i dont
understand why the wave cycles vary with this control setting.
 
Physics news on Phys.org
You questions are not very clear and are therefore had to answer.
Here is a java applet for lissajous lab. It did not work with my browser but maybe it will for yours.
http://www.math.com/students/wonders/lissajous/lissajous.html
here is one that worked on my browser.
http://abc.net.au/science/holo/liss.htm

The oscilloscope takes some getting use to. Play with it a little and don’t be afraid of breaking it, most are rather resilient. I have only seen people mess up oscilloscopes when they mess with the output impendence of the oscilloscope.
 
The horizontal time setting (time/div) controls the time it take the sweep to cross the screen. It is specified as the time required to travel 1 major division on the screen. If you are displaying periodic signal the number of cycles on the screen will be determined by the horizontal time setting. So if you are looking at a 60hz it has a period of ~17ms. So a setting of 10ms/div will mean that 1 cycle of the signal will occur in about 1.7 divisions. If there are 10 divisions across the screen you will see about 5 cycles. Now if you change the time base to 5ms/division each cycle will take about 3 divisions and only ~3 cycles will be displayed.

To create Lissajous patterns you must provide a signal to drive the horizontal sweep. This means that your horizontal speed controls are disabled. The normal signal applied to the horizontal deflection is a sawtooth, this causes the sweep to cross the screen at the uniform rate specified by the time base. When you apply a sinusoidal signal to the horizontal plates the rate the sweep travels is no longer linear, it now varies with the phase of the input signal. When you apply equal frequencies to both horizontal and vertical plates the result is the circle. The appearance of the pattern is determined by the ration of the horizontal to vertical frequency. A figure 8 means a 2:1 ratio, a 1:2 yields a \infty Higher ratios yield more interesting results.

The key to understanding the Lissajous pattern is realize that the horizontal sweep rate is not linear.
 
Last edited:
figs said:
i'm learning about oscilloscopes, and i don't understand
why this Lissajous figure can be go from a straight line to
a circle.

If your two inputs are in-phase, they will produce a straight line, if they are 90 degrees out of phase, they will produce a circle (provided the two inputs are the same frequency of course).

Claude.
 
Claude Bile said:
If your two inputs are in-phase, they will produce a straight line, if they are 90 degrees out of phase, they will produce a circle (provided the two inputs are the same frequency of course).

Claude.
Yes, reflecting on my post, I realized that I had neglected to mention the importance of the phase relationship between the signals. Thank you Claude.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top