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P.H.Y.S.I.C.S P2 D.O.U.B.T.S CIE

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halosh92:

--- Quote from: The SMA on June 08, 2010, 11:18:46 am ---m/j 2006 p2 Q6b), why do we need to times 2 the wavelength, using v=f x lambda
and for part a) how to draw the waves?

--- End quote ---

ok the wavelength they gave is basically the length between one antinode and another successive antinode ok?
(two positions loud voice as they mentioned)
so...
if u draw the wave  u will first have an antonide then node then antinode the normal wave
so the distance from one antinode to the node is 32.4/2 ok?
always remember distance from node to node = half a wavelength
and wavelength from node to antinode= quarter a wavelength
so wat we found is quarter a wavelength now multiply by 4
got it?

halosh92:
could someone briefly explain systematic and random errors????

Chingoo:

--- Quote from: halosh92 on June 08, 2010, 11:38:56 am ---ok the wavelength they gave is basically the length between one antinode and another successive antinode ok?
(two positions loud voice as they mentioned)
so...
if u draw the wave  u will first have an antonide then node then antinode the normal wave
so the distance from one antinode to the node is 32.4/2 ok?
always remember distance from node to node = half a wavelength
and wavelength from node to antinode= quarter a wavelength
so wat we found is quarter a wavelength now multiply by 4
got it?

--- End quote ---

I think you kinda lengthened the whole story...just to add my input (in case it helps), the distance between two consecutive nodes or two consecutive antinodes is lamda/2. See this maxima and minima as a waveform with displacement (y-axis) against distance (x-axis); the minima is the node and maxima the antinode. The antinode can be on either side of the x-axis, because if you remember displacement is taken from a reference point. In case of waves the reference point is the mean position (the node) and the maximum displacement from this mean position is an antinode. If the antinode is a point 2 cm from the mean position, the least displacement is NOT -2 but 0. Hence, like halosh said, simply multiply 32.4 by 2 and you have the lamda.

In times of a sine wave, a node and antinode are 90 degrees out of phase with each other (1/4 of the sine wave graph) and an antinode and another are (270-90=)180 degrees out of phase with each other (1.2 of the sine wave graph).

halosh92:

--- Quote from: Chingoo on June 08, 2010, 12:10:33 pm ---I think you kinda lengthened the whole story...just to add my input (in case it helps), the distance between two consecutive nodes or two consecutive antinodes is lamda/2. See this maxima and minima as a waveform with displacement (y-axis) against distance (x-axis); the minima is the node and maxima the antinode. The antinode can be on either side of the x-axis, because if you remember displacement is taken from a reference point. In case of waves the reference point is the mean position (the node) and the maximum displacement from this mean position is an antinode. If the antinode is a point 2 cm from the mean position, the least displacement is NOT -2 but 0. Hence, like halosh said, simply multiply 32.4 by 2 and you have the lamda.

In times of a sine wave, a node and antinode are 90 degrees out of phase with each other (1/4 of the sine wave graph) and an antinode and another are (270-90=)180 degrees out of phase with each other (1.2 of the sine wave graph).

--- End quote ---


hahah yes my mother always said i talk alot  ;D

cashem'up:
hey guys one question i had.... if i kept a filament in constant temperature lets say in Ice then would its IV graph be a straight line through origin

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