Month: September 2016

137 Lissajous Figures on CRO

When the CRO is set to the xy mode, it uses the Channel 1 input as the time-base. This means the x and y position of the CRO trace is controlled by inputs from Channel 1 and 2 respectively.

If both channels are fed with sinusoids with the same frequency, then the following traces wil result, depending on the phase relationship between the two sinusoids.

Picture1

In-phase. Much like the relationship between force and acceleration.

Picture2

Anti-phase. Much like the relationship between acceleration and displacement for a SHM.

Picture3

A quarter-cycle out of phase. Much like velocity leading displacement for a SHM.

In this demo, because I was using analogue signal generators, I was unable to set the inputs to the exact frequencies I wanted. So even though I was aiming for 100 Hz for both channels, I could only get them to be close to but not exactly 100 Hz. This means there was a slight difference in the periods of the two signals. This causes their phase relationship to keep shifting, alternating between in-phase and anti-phase. This results in the lovely outcome of the tracing alternating between diagonals and flipped diagonals, including the ellipses between.

When Channels 1 and 2’s frequencies are a nice integer ratio of each other, very intricate patterns are formed. By counting the number of times the trace cuts each axis, one can figure out the exact ratios. If you are interested and want to read up more, just google “Lissajous Figures”.

136 Ball tied to String in Circular Motion

The tension in the string is changing during the circular motion. Why?

Firstly, note that the vertical circular motion of a ball swung on a string is not uniform circular motion where the speed is constant.Gravity causes the ball’s speed to be fastest at the bottom and slowest at the top. (vB>vT) As such the required centripetal force is maximum at the top of the circle, and minimum at the bottom of the circle. (mv2/r)

a

There are thus two reasons why the string is most taut at the bottom.

  1. The required centripetal force at the bottom is maximum.(mvB2/r)
  2. Weight of the ball is in opposite direction to the centripetal direction. The tension in the string must overcome the weight before contributing to the required centripetal force. (T = mvB2/r + mg)

b

There are also two reasons why the string is least taut at the top.

  1. The required centripetal force at the top is minimum. (mvT2/r)
  2. Weight of the ball is in the centripetal direction. The weight assists the string to provide the required centripetal force. (T = mvT2/r  mg)

In fact, if the ball is swung too slowly, gravity may pull the ball down so quickly that the string becomes slack and the ball becomes a projectile motion instead. The minimum launch speed in order to complete the circular motion successfully is presented in this xmtutor video.