Prelim Physics Thread (2 Viewers)

Green Yoda · Jun 17, 2016

calamebe said:
My teacher gave me a sheet today for which we have to graph displacement-position graphs of waves. I'm kinda confused, as displacement and position are the same thing, at least going off the motion topic. So is there any kind of difference in the definition of these terms for waves, or is it most likely a typo (I find that a bit unlikely though as it has it many times)

It doesn't sound right..as there will be two dependent variables.

Suu · Jun 17, 2016

calamebe said:
My teacher gave me a sheet today for which we have to graph displacement-position graphs of waves. I'm kinda confused, as displacement and position are the same thing, at least going off the motion topic. So is there any kind of difference in the definition of these terms for waves, or is it most likely a typo (I find that a bit unlikely though as it has it many times)

The question is asking you to graph the wave as a function of position, aka how far it has traveled from its original position (in the direction of its propagation). So, it is asking you to plot basically the amplitude of the wave against the distance from its source. (it should look like a sine wave, and from it you should be able to find its wavelength and amplitude.

Suu · Jun 17, 2016

InteGrand said:
From a quick glance, Q.22 to Q.30 are by definition.

I would argue 25 is by definition as well.

InteGrand · Jun 17, 2016

Suu said:
I would argue 25 is by definition as well.

Yeah I meant all the Q's from 22 to 30 (so the ones in between too).

Suu · Jun 17, 2016

InteGrand said:
Yeah I meant all the Q's from 22 to 30 (so the ones in between too).

I missed the "to" in between, sorry bout that heh

eyeseeyou · Jul 4, 2016

I'm having trouble visualising this question

In order to be readable, the intensity of the light from a slide projector showing the words of a song needs to be 900Wm^-2. When thee projector showing the words of a song needs to be 900Wm^-2. When the projector was placed 5.00 from the screen, the intensity was 1600Wm^-2. What is the maximum distance the projector can be found from the screen so that the words are still readable?

Thanks

eyeseeyou · Jul 4, 2016

Also I am struggling to understand one other question

Cathode ray tubes are used to display a person's heartbeat in hospital. The display consists of a series of complex waves on the screen of a cathode ray oscilloscope. Another monitor records the pulse rate. The diagram represents one full tracing of a person's heartbeats with a rate of 72 beats per minute. What is the speed of the trace as it was moved across the screen, in this case 18cm wide?

Once again thanks

InteGrand · Jul 4, 2016

eyeseeyou said:
I'm having trouble visualising this question

In order to be readable, the intensity of the light from a slide projector showing the words of a song needs to be 900Wm^-2. When thee projector showing the words of a song needs to be 900Wm^-2. When the projector was placed 5.00 from the screen, the intensity was 1600Wm^-2. What is the maximum distance the projector can be found from the screen so that the words are still readable?

Thanks

$\noindent Have you tried using inverse square law? According to this law,$

$\noindent $I \equiv I\left(d\right) \propto \frac{1}{d^2}$,$

$\noindent where $I$ is the intensity in ($\text{W m}^{-2}$ here) and $d$ is the distance (in metres here).$

$\noindent We are given that $I(5) = 1600$. We want the $d$ such that $I\left(d\right) = 900$, as at this $d$, the words are just readable, and any further than this, the intensity decreases to below $900$ and so the words are no longer readable. So using the proportionality condition, we have $\frac{I\left(d\right)}{I\left(5\right)}=\frac{5^2}{d^2}\Rightarrow \frac{900}{1600}=\frac{5^2}{d^2}\Rightarrow d^2 = \frac{5^2 \times 1600}{900}\Rightarrow d = \frac{5\times 40}{30}=\frac{20}{3} $. So the answer is $\frac{20}{3}$ metres, which is $6.67$ m to three significant figures.$

eyeseeyou · Jul 4, 2016

InteGrand said:
$\noindent Have you tried using inverse square law? According to this law,$

$\noindent $I \equiv I\left(d\right) \propto \frac{1}{d^2}$,$

$\noindent where $I$ is the intensity in ($\text{W m}^{-2}$ here) and $d$ is the distance (in metres here).$

$\noindent We are given that $I(5) = 1600$. We want the $d$ such that $I\left(d\right) = 900$, as at this $d$, the words are just readable, and any further than this, the intensity decreases to below $900$ and so the words are no longer readable. So using the proportionality condition, we have $\frac{I\left(d\right)}{I\left(5\right)}=\frac{5^2}{d^2}\Rightarrow \frac{900}{1600}=\frac{5^2}{d^2}\Rightarrow d^2 = \frac{5^2 \times 1600}{900}\Rightarrow d = \frac{5\times 40}{30}=\frac{20}{3} $. So the answer is $\frac{20}{3}$ metres, which is $6.67$ m to three significant figures.$

Yup and it doesn't work out for me

Thanks

InteGrand · Jul 4, 2016

eyeseeyou said:
Yup and it doesn't work out for me

Thanks

Which doesn't work out?

eyeseeyou · Jul 4, 2016

InteGrand said:
Which doesn't work out?

When I'm trying the inverse square law but dw bc you helped me with it

Green Yoda · Jul 4, 2016

eyeseeyou said:
Also I am struggling to understand one other question

Cathode ray tubes are used to display a person's heartbeat in hospital. The display consists of a series of complex waves on the screen of a cathode ray oscilloscope. Another monitor records the pulse rate. The diagram represents one full tracing of a person's heartbeats with a rate of 72 beats per minute. What is the speed of the trace as it was moved across the screen, in this case 18cm wide?

Once again thanks

f=1/T
f=1/72
v=fλ
v=1/72*0.18
v=0.0025
v=2.5x10^-3

eyeseeyou · Jul 4, 2016

Rathin said:
f=1/T
f=1/72
v=fλ
v=1/72*0.18
v=0.0025
v=2.5x10^-3

Thanks

Green Yoda · Jul 4, 2016

I made mistake in my previous post ^
T is not = 72
f=72/60 =1.2
so v=1.2*0.18
v=0.216m/s

eyeseeyou · Jul 4, 2016

Rathin said:
I made mistake in my previous post ^
T is not = 72
f=72/60 =1.2
so v=1.2*0.18
v=0.216m/s

Once again thanks

Simorgh · Jul 7, 2016

How do you identity and find out the net displacement in this question?

Soz for poor image quality

Fizzy_Cyst · Jul 7, 2016

Simorgh said:
How do you identity and find out the net displacement in this question?

Soz for poor image quality

Pythagoras twice

Simorgh · Jul 7, 2016

Fizzy_Cyst said:
Pythagoras twice

Care to elaborate further? Do you do Pythagoras twice to find B, but how do you know the angle between the side 250m and side b is perpendicular for that to work?

Where is the net displacement in this question? If I know that then I could do this easily.

Soz kinda confused.

Fizzy_Cyst · Jul 7, 2016

Simorgh said:
Care to elaborate further? Do you do Pythagoras twice to find B, but how do you know the angle between the side 250m and side b is perpendicular for that to work?

Where is the net displacement in this question? If I know that then I could do this easily.

Soz kinda confused.

Pythagoras once to find 'a' -- it seems you have done this

Second Pythagoras comes between 'a' and the vertical displacement of 250m.

'a' vector would lie on the plane which contains the cardinal points (I.e., parallel to earth surface or ) and 'up' is perpendicular to earths surface, so these two form aright angled triangle also

That second Pythagoras will give you the magnitude of the net displacement

Then you would need to determine the angles. You would need an angle with respect to cardinal points Aswell as an angle of elevation as it is a 3D question (beyond the syllabus really)!

Pretty tedious question.

jathu123 · Jul 22, 2016

please help, an explanation if possible. Thanks

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