Prelim Physics Thread (4 Viewers)

Green Yoda · Apr 30, 2016

InteGrand said:
For 1., if we first double the power of the source, then the intensity doubles. Then if we double the distance from the source, this has the effect of 'quartering' the intensity. Since 2*(1/4) = 1/2, the end result is that the intensity is halved.

I still dont understand..
so 2I=1/(2)^2
therefore 2I=1/4
then I would be 1/8?

InteGrand · Apr 30, 2016

Rathin said:
I still dont understand..
so 2I=1/(2)^2
therefore 2I=1/4
then I would be 1/2?

Are you saying the final intensity is half the original one? That is the answer that I gave.

Green Yoda · Apr 30, 2016

InteGrand said:
Are you saying the final intensity is half the original one? That is the answer that I gave.

sorry typo there I edited it..

Nailgun · Apr 30, 2016

Rathin said:
I still dont understand..
so 2I=1/(2)^2
therefore 2I=1/4
then I would be 1/2?

Basically, the factor of which the intensity will change is the inverse square of the factor change in distance

Let original intensity be I
Intensity is doubled so it will become 2I

Factor of distance change is 2 (doubled)
inverse is 1/2
inverse squared is 1/4

2I * 1/4

= 1/2 * I

so intensity is halved

Green Yoda · Apr 30, 2016

Nailgun said:
Basically, the factor of which the intensity will change is the inverse square of the factor change in distance

Let original intensity be I
Intensity is doubled so it will become 2I

Factor of distance change is 2 (doubled)
inverse is 1/2
inverse squared is 1/4

2I * 1/4

= 1/2 * I

so intensity is halved

if 2I=1/4..then when you divide by 2..I becomes 1/8?

InteGrand · Apr 30, 2016

Rathin said:
sorry typo there I edited it..

$\noindent What do you mean by $2I = \frac{1}{2^2}$? We should carefully define our variables. Here is one way to do it.$

$\noindent Let the original intensity be $I_0$ and the new one be $I_1$ (i.e. after doubling the distance and doubling the power of the source). Let $I^{\star}$ be the intermediary intensity (i.e. after we have only doubled the power of the source say). Then since intensity is proportional to source power, we have $I^{\star}=2I_{0}$. Now, $I_1$ is the intensity after now doubling the distance too. By the inverse square law, it follows that $I_{1} = \frac{1}{4}I^{\star}=\frac{1}{4}\times 2 I_{0} = \frac{1}{2}I_{0}\Rightarrow$ the new intensity ($I_1$) is half the original one ($I_0$).$

The above can also be done by inspection (which is how I did in my original post).

Nailgun · Apr 30, 2016

Rathin said:
if 2I=1/4..then when you divide by 2..I becomes 1/8?

nah the final intensity is equal to 2I * 1/4

okay maybe a formula will help you

F means final, i means initial
d is distance, I is intensity

${ I }_{ F }=I_{ i }\quad \times \quad \left( \frac { { d }_{ F } }{ { d }_{ i } } \right) ^{ -2 }$

Nailgun · Apr 30, 2016

Also here you go

${ I }_{ F }=I_{ i }\quad \times \quad \left( \frac { { d }_{ F } }{ { d }_{ i } } \right) ^{ -2 }\\ \\ Let\quad the\quad intensity\quad before\quad any\quad changes\quad be\quad c\\ \\ I_{ i }=2c\quad (as\quad the\quad intensity\quad is\quad doubled)\\ \\ \frac { { d }_{ F } }{ { d }_{ i } } =2\quad (as\quad it\quad is\quad doubled)\\ \\ { I }_{ F }=2c\quad \times \quad \left( 2 \right) ^{ -2 }\\ \\ \\ { I }_{ F }=2c\quad \times \quad \frac { 1 }{ 4 } \\ \\ { I }_{ F }=\frac { 1 }{ 2 } c\\ \\ Hence,\quad the\quad intensity\quad has\quad halved\quad from\quad the\quad beginning\\$

Green Yoda · Apr 30, 2016

I think I have gotten the concept wrong..
d=1, I=1
d=1/(2)^2 = 2I
and so on

but In actual its
d=1, I=1
d=2, I=1/(2)^2

So when I visualise it at 2d the intensity is 1/4 and if you double that it becomes 1/2

Nailgun · Apr 30, 2016

Rathin said:
I think I have gotten the concept wrong..
d=1, I=1
d=1/(2)^2 = 2I
and so on

but In actual its
d=1, I=1
d=2, I=1/(2)^2

So when I visualise it at 2d the intensity is 1/4 and if you double that it becomes 1/2

yeah pree much, although we doubled it first and then did it, but I think it's fine either way

basically as the distance increases, the intensity of the light decreases at an increasing rate

InteGrand · Apr 30, 2016

It's easier to just think about it intuitively. It's obvious from an intuitive point of view that increasing the distance will lead to intensity going to down (e.g. if you take a heat source away from you, you feel less heat) and increasing the power of the source leads to intensity going up (e.g. if you have a heater near you and increase it's power, you'll feel more heat). So all you need to know now is how the factors work.

Just know that if you increase the distance by some factor, the intensity goes down by the square of the factor that the distance increased by (this is the inverse square law said in words). Also, if you increase the source's power by a factor, the intensity goes up by the same factor.

So for instance, if you double a distance, the intensity will go down by a factor of 2² = 4. Then if you double the source's power, the intensity compared to before doubling the power will be increased by a factor of 2.

A combination of increasing by a factor of 2 and decreasing by a factor of 4 is an overall impact of decrease by a factor of 2, i.e. halving (because if you double a number and then quarter it, overall it's halved).

Green Yoda · Apr 30, 2016

Yup need to remember that the Intensity is the outcome of the 1/d^2 and that is what's affected in terms of the 1/4, 1/9, 1/16....

Also, need urgent help on the question:
A sound travels through a tube fitted with various pressure gauges(high and low..so basically compressions and rarefactions inside the tube). Describe the distribution of air particles if a higher frequency travels through the tube.

Green Yoda · Apr 30, 2016

Rathin said:
Yup need to remember that the Intensity is the outcome of the 1/d^2 and that is what's affected in terms of the 1/4, 1/9, 1/16....

Also, need urgent help on the question:
A sound travels through a tube fitted with various pressure gauges(high and low..so basically compressions and rarefactions inside the tube). Describe the distribution of air particles if a higher frequency travels through the tube.

This is what I have wrote so far:
Inside the tube all the air particles will be compressed which will cause more particles to be under high pressure, creating a higher number of compressions and thus more rarefactions after every interval of compressions.

Would this be the correct and sufficient answer for the 2 marker question?

eyeseeyou · May 1, 2016

InteGrand said:
$\noindent What do you mean by $2I = \frac{1}{2^2}$? We should carefully define our variables. Here is one way to do it.$

$\noindent Let the original intensity be $I_0$ and the new one be $I_1$ (i.e. after doubling the distance and doubling the power of the source). Let $I^{\star}$ be the intermediary intensity (i.e. after we have only doubled the power of the source say). Then since intensity is proportional to source power, we have $I^{\star}=2I_{0}$. Now, $I_1$ is the intensity after now doubling the distance too. By the inverse square law, it follows that $I_{1} = \frac{1}{4}I^{\star}=\frac{1}{4}\times 2 I_{0} = \frac{1}{2}I_{0}\Rightarrow$ the new intensity ($I_0$) is half the original one ($I_0$).$

The above can also be done by inspection (which is how I did in my original post).

Why would we let one intensity be this and one intensity another thing?

InteGrand · May 1, 2016

eyeseeyou said:
Why would we let one intensity be this and one intensity another thing?

Because they represent two different things. If you use the same symbol for them, it's easy to get confused. It's like in a maths word problem if there are two unknown quantities and you let both of them be x.

eyeseeyou · May 15, 2016

How do ticker times relate to newton's second law?

eyeseeyou · May 15, 2016

Integrand...where r u?

Green Yoda · May 15, 2016

eyeseeyou said:
How do ticker times relate to newton's second law?

http://practicalphysics.org/relationships-between-acceleration-force-and-mass.html

eyeseeyou · May 15, 2016

Rathin said:
http://practicalphysics.org/relationships-between-acceleration-force-and-mass.html

Thanks Rathin

You help me when Integrand wan't here

jathu123 · May 25, 2016

Question:
A driver of mass 75kg travelling in a 1200 kg sedan collides with a stationary police car with total mass of 1500 kg. The tangled wreck moves off after the collision at 8 ms^-1. Assuming negligible frictional force on both vehicles: Calculate the impulse on the police car due to the collision.

Just need confirmation cause I'm not 100% sure about the answer at the back
thankyou very much

Prelim Physics Thread (4 Viewers)

Hi Φ

Well-Known Member

Hi Φ

Cole World

Hi Φ

Well-Known Member

Cole World

Cole World

Hi Φ

Cole World

Well-Known Member

Hi Φ

Hi Φ

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Hi Φ

Well-Known Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 4)