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A2 Physics | Post your doubts here

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can any1 plz tell me wen to use cos and wen to use sine in any kind of oscillation...
it depends on the graph... usually if you're taking x as the midpoint of the oscillation when t=0, then it's a sine graph. If you're taking x as the maximum amplitude position when t=0, its a cosine graph.

Think about it, t is on the x axis. If t=0, sin(0)=0 and cos(0)=1 so if the formula is x=x,sin(wt) then at t=0, x=0 which is the equilibrium position.. and if the formula being used is x=x,cos(wt) then x=x, at t=0 which is the maximum displacement.

x= displacement
x,= max displacement
w= omega- angular frequency
t= time
 
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can any1 plz tell me wat is meant by Analoge signal and Digital signal?
Digital Signal is the one in which there is a series of pulses btw descrete levels and in Analogue, the waveform of signal is in a similar manner as to the waveform of sound wave producing it
 
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(A)Electrons in a cathode-ray tube leave the cathode with negligible speed at a potential of -9000V and are accelerated to an anode at a potential of -200V. For an electron in this tube calculate
(i) the gain in electrical energy
(ii)the loss in potential energy
(iii)the gain in kinetic energy
(iv)the speed on reaching the anode
(B) Explain why (A)(i) is a gain but (A)(ii) is a loss.
 
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Digital Signal is the one in which there is a series of pulses btw descrete levels and in Analogue, the waveform of signal is in a similar manner as to the waveform of sound wave producing it
what does this means that...digital has no intermediate values...
 
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(A)Electrons in a cathode-ray tube leave the cathode with negligible speed at a potential of -9000V and are accelerated to an anode at a potential of -200V. For an electron in this tube calculate
(i) the gain in electrical energy
(ii)the loss in potential energy
(iii)the gain in kinetic energy
(iv)the speed on reaching the anode
(B) Explain why (A)(i) is a gain but (A)(ii) is a loss.
ans..i..1/2QV...1/2*1.6*10^-19*(-200-(-9000))
so ans might be 7.04*10^-16...i guess...for the first 1
 
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only two values 1 and 0 . no intermediate value . no curve in the graph . a horizontal line at 1 or a horizontal line at 0
and wt does this means that...for analoge..signal has same varitaion (wit time) as the data...
 
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ans..i..1/2QV...1/2*1.6*10^-19*(-200-(-9000))
so ans might be 7.04*10^-16...i guess...for the first 1
yeah i did the same . and the gain in K.E would be equal to loss in P.E and we can find the speed by equating the K.E to 1/2mv^2 . Am i right?
 
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and wt does this means that...for analoge..signal has same varitaion (wit time) as the data...
it means that the signal has the same sketching or shape of graph as the data
e.g the voltage output would have the same variation with time(i.e the graph) as the input or whatever produced that voltage e.g sound .
 
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