Problem Solving Strategies

@yuvrajsingh I'm busy for a few moments ...

3:55 AM

@JohnRennie Good morning sir :-)

@JohnRennie I will wait sir

@yuvrajsingh These people have nothing else to do. They can work hard and earn money and buy what they love. If they are somehow caught in the act, the police will punish them along with GST from public :D :D

@user8718165 thank you,

@user8718165 time is not same, we all have to be aware from this antisocial elements, never live your home alone for more days,

Sorry it is leave not live

@yuvrajsingh yeah that's absolutely correct. I've come across local news where the hired guards conspire with the robbers.

4:12 AM

@user8718165 how the preparation is going on

@JohnRennie sir ping, when you get free

@yuvrajsingh I'm worried very much about CS for boards.my % will go down :-(

@yuvrajsingh sorry about this but a problem has happened at work. I have a dead server I have to fix. I shouldn't be too much longer.

Ok

4:37 AM

@yuvrajsingh hi, I'm back now.

@JohnRennie hi (it's kyle campbell) - I need some help with some SR/galilean transformation hw when you're available

@psa hi :-)

https://imgur.com/5RqaEU6
https://imgur.com/a/4HMn5AV
https://imgur.com/zbwg2uH

the third picture is the actual question

@psa reading now ...

@psa OK, that looks straightforward. What's the problem with the question?

I don't know why it's not x = x' + vt and x' = x - vt

4:47 AM

i and iv are both correct. Obviously so since it's the same equation rearranged.

ii and iii have the sign of the velocity wrong

It says I got it wrong

but I picked i and iv

Oh wait!

The velocity of the train in the frame x (the platform frame) is $-v$ because x increases to the right, so right is the positive direction, but the train is moving left.

Hmm, that's kind of ambiguous. Is $v$ a vector or is the magnitude of the velocity?

I think it's supposed to be a vector, from which we decide the sign relative to the frame we picked.

Well look at it this way. Suppose the origins coincide at time zero, then because the train is moving left as time goes on the origin of the x' frame will be at a negative position in the x frame. So $x = x' - |v|t$

ah, okay

4:57 AM

Where I've written $|v|$ to make it clear I mean the magnitude of the velocity.

I always have to be careful with the signs I find...

The question is ambiguous.

The correct expression is $x = x' + vt$, but $v$ can be a positive or a negative number depending on whether thetrain is moving right or left.

right

but moving to the left ---> $v<0,$ but I see the ambiguity there because if you put - vt if $v<0$ then -vt = +vt

imgur.com/z13Kt4c that's the next one I'm having issues with

Yes, if the train is moving left the velocity is still just $v$ but $v = -|v|$

@psa aha, this time they specifically tell you that $u > 0$

ha yeah

5:05 AM

I don't think it's that complicated. Frame x'' just has an additional transformation that reverses the sign of the distance. I would temporarily use a new variable $X'' = -x''$ then the transformation is simply $X'' = x - ut + A$ (the constant A just gives the relative positions at $t=0$).

Then just substitute for $X''$ giving $-x'' = x -vt + A$

Or $x'' = -x + vt + A$

hmm, I thought it would be x'' = -x' - vt + A but I guess I messed up the sign of v

Oh wait, damn, we have the interpretation of the sign issue again.

Well let's try a specific example. Suppose the origins coincide at $t=0$ so $A=0$. Now take $t=1$ so the origin of x'' is at $x=u$ in the x frame (we know u is positive).

My equation is $x'' = -x + vt$ so that's $x'' = -(+u) + (+u) = 0$ and that works because we're finding the origin in x''.

right

actually, I guess the +vt might make sense

because in the x'' frame, the x origin is heading toward the left

er, i mean towards the right

agh, that's where I'm confused - I thought it should be -vt because in the x'' frame, the x origin is moving "to the right," which is negative in the x'' frame

5:27 AM

ah, no I see

It is a bit of a mess ...

@JohnRennie If you're trying to calculate the time dilation for a clock moving at a speed v (t') relative to the rest frame of the clock (t), the formula is just $t' = {\gamma}t$ right?

we derived this with a ball bouncing between two plates and some trig

@psa you need to be very careful about just throwing factors of $\gamma$ around because it can lead you astray.

I strongly advise that when you are doing a calculation in SR you define the events of interest then use the Lorentz transformations to find those events in the other frame.

5:43 AM

yeah... because I think for this problem here:https://imgur.com/Bv6naYr
to calculate Aidan's time of his round trip, it would actually be (time of his round trip)/$\gamma$

because the clock's rest frame is actually in his frame

Suppose the travellers depart Earth at time zero, so in all frames the departure event is $(t=0, x=0)$.

In the Earth frame our traveller departs at a speed $v$ and travels a distance $d$ so in the Earth frame the traveller arrives at the star at the event $(t = d/v, x = d)$. OK so far?

yes

So we want to find the time in the travellers frame when they reach the destination.

And the transformation for this is:

$$ t' = \gamma\left( t -\frac{vx}{c^2}\right) $$

Plugging in our values for x and t we get:

$$ t' = \gamma\left( \frac{d}{v} -\frac{vd}{c^2}\right) $$

This initially looks a bit intractable, but what we do is take a factor of $d/v$ out of the bracket to get:

$$ t' = \gamma \frac{d}{v} \left( 1 -\frac{v^2}{c^2}\right) $$

@psa OK so far?

yes

Because $d/v = t$ and $(1 - v^2/c^2) = 1/\gamma^2$

So this simplifies to:

$$ t' = \frac{t}{\gamma} $$

Which is of course what you said right at the outset :-)

But we've arrived at it in a methodical way that will always work even in more complicated situations.

5:57 AM

very nice

In this particular question we want $t'$ to be the same for Aidan and Nadia.

So they stay the same age.

(we're only treating half the journey, but the journey is symmetrical so the return half takes exactly the same time as the outward half)

yes

So we want $t'_A = t'_N$

We'll write $t_A = d_A/v_A$, and the same for Nadia, so we equate the two $t'$ values to get:

$$ \frac{d_A}{v_A\gamma_A} = \frac{d_N}{v_N\gamma_N} $$

where did the factor of $\gamma$ come from?

And then you solve for $v_N$

6:01 AM

same as before?

From our work above we figured out that:

$$ t' = \frac{t}{\gamma} $$

So for Aidan $t' = t_A/\gamma_A = (d_A/v_A)/\gamma_A$

And likewise for Nadia

right

Actually, wait, have I done the last part of this correctly?

The result $t' = t/\gamma$ is certainly correct.

I need to work now anyhow. I'll have to get back to you.

I think so yes

cheers!

Martin Sleziak

6:24 AM

@yuvrajsingh You can see the list of all room owners in the room info. Any moderator has all rights that room owners do. Relevant Meta Stack Exchange post: Is there a list of SE chat privileges, and the minimum reputation required for those privileges?

7:06 AM

@JohnRennie I think it would make sense that she would have to be travelling faster in order to be the same age as him

they won't necessarily be on Earth at the same time, but once they both are, she would have to travel faster in order for their clocks to be synchronized

@psa I was just thinking that it's a bit more complicated than just $t'_A = t'_N$ ...

7:26 AM

@JohnRennie hu

@yuvrajsingh hi

@JohnRennie yes, I ended up doing it differently

that result gave me a non-real root for v

@JohnRennie what is difference in term of energy in standing and transverse wave

I have another problem... I'm working in a lab right now, and I'm trying to design an RF coil to basically kick hot atoms (at the end of the Maxwell-Boltzmann distribution) out of a cold atom trap. I found a result that says that the inductance goes like the square of the number of turns N, but where does this come from?

I read in standing wave particle oscillate at their position and in transverse wave it transmit vibration to the near particle

@JohnRennie I read in standing wave particle oscillate at their position and in transverse wave it transmit vibration to the near particle

7:29 AM

@yuvrajsingh a standing wave is equivalent to two waves of equal amplitude moving in opposite directions. Each wave transports energy in the direction it moves, so because we have equal and opposite waves the net energy transport by a standing wave is zero.

@JohnRennie what about the particles

@yuvrajsingh A single wave transports energy, so the energy flows from a particle to the particle next to it. This is what your book means by transmit vibration to the near particle.

In a standing wave we have two waves moving in opposite directions, so any single particle receives energy from the two particles on either side of it.

@psa you need to check out the derivation for the inductance of a coil. You do indeed end up with an $N^2$ term.

I must have done something wrong then, because I started with the Maxwell-Ampere law and ended up with it scaling like N. I can't find anything on google.

$L = N\frac{\phi}{I}$

The equation for $\Phi$ also contains a factor of $N$. That's where you get the $N^2$.

Oh, I see... $\phi = BA = \frac{{\mu_{0}}N}{I\ell}$?

7:38 AM

Yes, exactly.

OK, I see

@JohnRennie opposite wave has same amplitude as coming wave in standing wave

@yuvrajsingh yes

@JohnRennie there should be same phase mean going wave has phase pi then coming wave has phase - pi

There is superposition principle while wave combine somewhere constructive and some where destructive

@yuvrajsingh it's more complicated than that because these are travelling waves so they have the form $y(t,x) = A\sin(\omega t - kx)$

For the wave moving to negative $x$ we have $y(t,x) = A\sin(\omega t + kx)$ so the total wave is:

$$ y(t,x) = A(\sin(\omega t - kx) + \sin(\omega t + kx)) $$

7:53 AM

@JohnRennie this can be zero or negative or positive

@yuvrajsingh you mean y(t,x) can be zero or negative or positive? If so then yes.

Yes

But what you find is that there are points where y(t,x) is always zero. These are the nodes in the standing wave.

@JohnRennie is there any difference in standing wave and transverse wave at any point

In term of equation

You mean consider fixed x and look at the variation of y with time?

7:56 AM

Yes sir

Well consider the point x = 0.

For a travelling wave the equation is y(t,x) = A sin(wt - kx) so at x = 0 y(t) = A sin(wt)

For a standing wave y(t,x) = A(sin(wt-kx) + sin(wt-kx)) so at x = 0

y(t) = 2A sin(wt)

Amplitude double hmm~

So in both cases if we sit at fixed x and just watch y we get a sinusoidal motion.

@yuvrajsingh because there are two waves.

@JohnRennie what decides no of nodes or antinodes s in standing wave, and nodes where amplitude of wave is zero but there are several antinodes where amplitude is non zero

@JohnRennie I mean this i.stack.imgur.com/uMuTo.png

@JohnRennie, Could you please explain why systematic motion of a container filled with ideal gas doesn't contribute to its temperature? For example, temperatures of two identical containers with ideal gases, one at rest and the other travelling at a velocity of 7000 km/s, are equal. Why this is so?

We know that temperature on the absolute scale is directly proportional to the velocity of the molecules squared. But here its contradictory.

Kindly reply when you find time.

8:06 AM

Answer is avg k. E

For ideal gase 2/3RT @Intellex

Ok @yuvrajsingh, But average KE is frame dependant right?

@Intellex the temperature is proportional to the RMS velocity of the gas molecules in the rest frame of the gas.

i.e. suppose <v> is the average velocity of the gas then we take the root mean square value of v - <v>.

@Intellex The definition of temperature in the kinetic theory of gases emerges from the notion of pressure. Fundamentally, the temperature of a gas comes from the amount, and the strength of the collisions between molecules or atoms of a gas.

The first step considers an (elastic) impact between two particles, and writes Δp=pi,x−pf,x=pi,x−(−pi,x)=2mvxΔp=pi,x−pf,x=pi,x−(−pi,x)=2mvx where the direction xx denotes the direction of the collision. This, of course, is considering that the two particles have opposing velocities before impact, which is equivalent to viewing the impact in the simple…

@Intellex read Unruh effect

Thank you very much for your replies @JohnRennie, @yuvrajsingh. But kinetic energy increases due to systematic motion of the container. Right?

Aladdin

@JohnRennie hi

8:12 AM

@yuvrajsingh Thanks, I read that, but in Wikipedia, its said its not rigorous

@Aladdin hi

Aladdin

U know about capacitor filter in rectifier

@JohnRennie please sir my last question

@yuvrajsingh there is a node every half wavelength, so the number of nodes and antinodes depends on the wavelength of the two waves. Your diagram shows three waves with different wavelengths.

@yuvrajsingh or were you asking why there are nodes at the ends?

@Aladdin yes

@JohnRennie yes why nodes at end

8:15 AM

@yuvrajsingh They are fixed right?

@JohnRennie it may be because of phase reversal of coming back, I mean in all three wave there is no reason at ends

@yuvrajsingh your diagram shows the sort of standing waves you get on a guitar string.

@JohnRennie yes sir

In that case the string is clamped at ends to the string cannot vibrate at the ends. That's why there are nodes there.

The fact the amplitude has to be zero at the ends means we get a phase change of pi when the string reflects so that the incident and reflected wave sum to zero.

@Intellex They are fixed right?
I didn't, t get what you are saying

8:19 AM

@yuvrajsingh I think he just means the same as I said.

i.e. the strings are clamped at the ends and can't move.

OK sir @JohnRennie one last why we always write wt-kx instead of wt+kx

@JohnRennie Hello !

For traveling wave going right

@Jasmine hi :-)

@yuvrajsingh we take right to be the direction of positive x, so a wave going right has to have positive velocity. OK so far?

Sorry for interrupting I will wait for the ongoing discussion to end,

8:24 AM

OK @JohnRennie sir I am done

@Jasmine restoring torque is double

X=rdtheta

Where r is l

3 mins ago, by Jasmine

The question asks to calculate time period of small oscillation. The spring constants have same spring constant k.

@Jasmine when I see multiple springs my immediate reaction is to use energy to calculate the total spring constant.

@JohnRennie I am little confused with that method

The time period is going to be $T = 2\pi \sqrt{\frac{k}{I}}$, where $k$ is the overall spring constant and $I$ is the moment of inertia. Yes?

The confusion is regarding Gravitation potential energy of rod

8:29 AM

@Jasmine is answer is root 24k/m

@Jasmine For small oscillations the vertical position of the rod centre of mass doesn't change, so you can ignore changes in the gravitational potential energy and just consider changes in the springs.

Sorry I was away from phone

@yuvrajsingh no

@JohnRennie yup

K overall constant or overall spring constant

Actually I'm now wondering if we can ignore the gravity. Let's ignore it for now and see what answer we get.

@JohnRennie Ohh

@JohnRennie I got a wrong answer with after doing that

Advil Sell

@Jasmine what's the answer is it something like 9/2 as a constant

8:39 AM

@Jasmine We're going to be considering angular motion i.e. if we write the angle of the bar to the vertical as $\theta$ we'll have a restoring force equal to $-k\theta$ for some constant $k$. It's that constant $k$ that's going to appear in our equation for the period.

@AdvilSell answer is $2π{\sqrt(4m/15k)}$

@JohnRennie sir if we do this by torsional torque and neglecting torque to weight then we k(ldtheta) l/2+k(l/2dtheta)l/2=ialpha is it right please look at it sir, because of them provide torque in same direction

@JohnRennie Ok

@Jasmine I'm just drawing a diagram ...

@JohnRennie Ok

8:47 AM

@Jasmine Are we told the spring constants?

Both springs have the same constant $k$?

@JohnRennie yes

OK so the extension of the lower spring is $x = L/2 \sin\theta \approx \frac{L}{2} \theta$

@JohnRennie yes

and the extension of the upper spring is $L\theta$.

@JohnRennie yes

8:52 AM

So the force due to the lower spring is $F = k\frac{L}{2}\theta$ and the torque is $FL/2 = k(\frac{L}{2})^2\theta$

@JohnRennie yes

Likewise the force due to upper spring is $F = kL\theta$ so the torque is $kL^2\theta$.

So if we ignore gravity for now the total torque is $k\frac{5L^2}{4}\theta$

@JohnRennie I have written same only

And the moment of inertia of a rod about its end is $I = mL^2/3$

So our equation is going to be:

$$ \ddot \theta = \frac{\tau}{I} = k\frac{15}{4m}\theta $$

@JohnRennie yup the answer is matching

But its not obvious to me why they have ignored gravity

8:58 AM

Yes, there would be a gravitational force of $mg\sin\theta$, which is approximately $mg\theta$. And this force acts in the opposite direction to the springs.

It was taught that if someway the gravity is helping to balance the spring forces ( initially ) then we dont consider gravity while writing final torque

But here initially the springs are at natural length but gravity is only balancing hinge force not spring force

So the total torque would be $\tau = (k\frac{5L^2}{4} - mg\frac{L}{2})\theta$

I can't see a good reason to ignore gravity, though including gravity gives a more complicated expression for the period. I guess the question assumes the springs are stiff enough that their force is far greater than the gravitational force.

@JohnRennie mgL/2 is not small compared to the k5L^2/4 term

In some questions gravity was considered

It could be. We aren't told the values of k and m. If $k \gg m$ the spring force would dominate.

I would just guess the answer is wrong

Advil Sell

9:04 AM

@Jasmine you can't says that unless given

I agree it's a poor question.

@AdvilSell yup but generally its not so small

@JohnRennie while following the energy method , is the gravitational potential energy we write, considering reference line as equilibrium position

@Jasmine Yes. The decrease in height at an angle $\theta$ would be $h = L/2 - L/2 \cos\theta$

@JohnRennie Ok

So we write that as $h = L/2(1 - \cos\theta)$ and $\cos\theta \approx 1 - \theta^2/2$

9:09 AM

@JohnRennie to the reference line is at the hinge or centre of mass

When the rod is vertical the height of the COM is $L/2$

And when the rod is tilted at an angle $\theta$ the height of the COM is $L/2\cos\theta$

So the decrease in height at an angle $\theta$ is $h = L/2 - l?2\cos\theta$

@JohnRennie Why is it wrong to consider zero GPE at hinge

@Jasmine we want the change in the GPE with angle

It doesn't matter where you take the origin. The change in the GPE will be the same.

@JohnRennie yess.. silly on my part

Got it Thanks :-)

@Jasmine Cool :-)

9:47 AM

@JohnRennie what is normal. temperature at Chester, is there any difference in weather of Chester and london

@yuvrajsingh Chester is in the west of the UK while London is in the east.

As a general rule the west is warm and wet while the east is cold and dry.

But there isn't a huge difference in the temperatures. Maybe up to 5° C.

@JohnRennie what is temperature in my city

You know during summer days it goes around 50

Wow, that is hot. In Chester the temperature rarely gets above 30°C. At the moment it's 17°C.

@JohnRennie what most important this about my home state Is it s legacy, it is place who always face condition like drought

Hence our food very different from rest of India