@JohnRennie : as well as good at understanding the googled text.

@JohnRennie very good at understanding googled text.:p

@yuvrajsingh Hii Yuvraj :-)

Morning :) to all

@user8718165 hi

@JohnRennie hi

@yuvrajsingh morning

@Nobodyrecognizeable morning :-)

@JohnRennie ^^

The wires are behaving as springs, so the tension in the wire is proportional to the extension.

@JohnRennie OK.

Since it says $x \ll \ell$ I'm guessing they assume the angle $\alpha$ can be treated as constant.

@JohnRennie OK.

@JohnRennie Good morning sir :-)

If you pull the mass down a distance $x$ then the increase in the length of the wire is $d = x/\cos\alpha$. Yes?

@JohnRennie so we need to find effective spring constant? Ie two springs are parallel?

@JohnRennie yes.

@Nobodyrecognizeable and the tension in the wires is $T = Y \frac{d}{\ell}$

@JohnRennie yes.

And the vertical force on the mass (from one wire) is $F = T\cos\alpha$

@JohnRennie yes. Ie from two will be $F_{res} =2Tcos\alpha$

Double that and you get the total force as a function of $x$, and that gives you the spring constant. Then the time constant is $2\pi\sqrt{m/k}$

@JohnRennie OK.

Although I'm not sure I understand why $r$ appears in the formula.

Oh, of course, it appears in the equation for the tension in the wires.

$Y\frac{x}{\ell}$ gives you the stress i.e. the force per unit area, so you have to multiply it by $\pi r^2$ to get the force.

I misread the question and didn't realise $r$ is the radius of the wire.

@JohnRennie hi

@yuvrajsingh morning :-)

I am studying about sound pressure wave

Yes?

@JohnRennie it doesn't give me any of the answers.

I have a small query @JohnRennie

@Nobodyrecognizeable Give me a monent and I'll have a go at the calculation

@JohnRennie OK.

Let say I have a material x and dx (any time time t partical at x suffer a displacement dx), as the wave passes the ends x and dx are shifted by around s and s+ds, what will be the increase in volume. @JohnRennie

@JohnRennie sorry I got c.

@Nobodyrecognizeable you got there just before me. I was about to say I get (c) :-)

@JohnRennie have a nice day professor and of course thanks for the help.

Sorry for the trouble

I wrote delta v=AdeltaS@JohnRennie

@yuvrajsingh Can you clarify what you mean by $x$ and $dx$? Presumably $x$ is the horizontal distance, so $dx$ is the displacement due to the sound wave at the distance $x$ i.e. we'd have an equation like $dx = A\sin(\omega t - kx)$.

Yes @JohnRennie

Actually if position x suffers displacement s then I wrote the equation s=sosin(wt-kx)

Actually I want to find what faction of volume changes of the point x and x+dx when they are displace s and ds @JohnRennie

Why the their is phase difference of pi/2 between pressure wave and longitudinal wave.

I have seen this derivation but I can't remember it ...

I'm afraid I need to work for a while now. Monday mornings are always busy for me.

@JohnRennie hi

@Aladdin hi

@JohnRennie hi sir

@yuvrajsingh hi. I'm still working I'm afraid.

There is question on stack exchange which comes in debate, can you please have look at my answer @JohnRennie

It is of three lines.

@yuvrajsingh link?

physics.stackexchange.com/questions/509039/… @JohnRennie

Your answer looks fine. If the forces weren't equal and opposite then as you say conservation of momentum would be violated.

, I have small question too on it. @JohnRennie

hello @JohnRennie

@JohnRennie ping when free

@user8718165 @Aladdin if JR SIR Is busy you can come in this chat room for your question owner Anna V mam

Here is the link. chat.stackexchange.com/rooms/100103/anna-v

@yuvrajsingh thank you very much for the link bro....but will she ever answer our questions as JR Sir does? ....BTW I'm happy to chat :-)

I haven,,t but, she is interested to teach the student like Jr sir do, without knowing her we can, t talk and make remark, I agree JR, no one can replace Jr but, our aim is to gain knowledge, upload the question, I believe she will. @user8718165

@JohnRennie Hi

@yuvrajsingh @user8718165 @Aladdin I have finally finished work!

@Aladdin hi

Sir @Aladdin first. @JohnRennie

don't worry ask.i am in codeclub

@JohnRennie hi sir

@yuvrajsingh yeah...agreed. I'll surely ask her questions. Please don't get me wrong. I was just worried if she'd get annoyed by our questions. If she's happy to chat then it's an extremely good news for us :-)

@JohnRennie hello Sir :-)

@yuvrajsingh sorry for the delay. What did you want to ask? About sound waves?

@JohnRennie hello sir...can I ask a qn ?

@user8718165 yes, Yuvraj doesn't seem to be around.

@user8718165 actually give me five minutes. I need to do something ...

I'm back ...

Aah sir are you there @JohnRennie

@yuvrajsingh hi :-)

Do you remember, I ask your comment on question @JohnRennie

That one?

Yes sir, while answering the question, I came across a small question in mind@JohnRennie

What's the question?

Sir I have a ball that strikes the wall according to Newton second whenever there is change in momentum body will experience the force ok

Yes

Why the wall will experience any force N during collision, it momentum haven't change during, after, and before it is zero, but then also during collision we have consider N normal reaction.

And if I accept then also I have

A doubt, suppose the big iron ball is any how pushing the wall, there also we consider Newton second law.

No change in momentum

But still there is normal reaction can you explain this

@JohnRennie

It's because you are assuming that the wall is fixed and cannot move.

So you have an external force keeping the wall in place.

That external force acts where the wall is attached to the Earth i.e. at the base of the wall.

In principle when you bounce a ball off the Earth both the ball and the Earth move after the collision, and this conserves the momentum. But in practice the Earth is so much heavier than the ball that we consider it fixed.

The price we pay for this is that with this assumption momentum is no longer conserved.

OK, when there is a big collision between two stars, did they move apart @JohnRennie

Because in space I can, t define

Something is in motion or not

@yuvrajsingh that's complicated because stars aren't solid objects. They are large balls of gas. So when stars collide they merge to form a single even larger ball of gas.

But if you collide two rubber balls in space they will bounce apart.

Why sir

How would you define a momentum of a body

Who itself depend upon the velocity

Which Is a frame dependent quantity @JohnRennie

Momentum is indeed a frame dependent quantity. When we talk about conservation of momentum we mean that if we define a frame and measure the momentum in that frame we find it is the same before and after the collisison.

For example suppose you are floating freely in space beside the colliding balls. You can measure the total momentum in your rest frame, and you'll find it remains constant.

Did floating mean rest, frame @JohnRennie

When I say floating I mean you are in inertial frame i.e. there are no external forces acting on you.

OK sir, I got your points, but in last do you have a example where momentum conservation fails. Because we assume it be conserve for small particle like proton and neutron. @JohnRennie

Momentum is always conserved. The only time it may appear that momentum is not conserved is if you are in a non-inertial frame.

I can rectify it by' relative velocity 'am I right sir. @JohnRennie

I'm not sure what you mean by that ...

I mean when I am moving, momentum conservation fails, can you explain it. @JohnRennie

Momentum conservation does not fail when you are moving, provided you are moving in a straight line at constant speed.

Let say I have two Ferrari moving towards each other, speed v and I am moving along Ferrari one with speed v+dv will the momentum be conserved. If no them how. @JohnRennie

Let's take a simpler example of two balls colliding on a table. Let's start with you standing besdie the table, and in this frame the balls have vlocities of $+v$ and $-v$ i.e. equal and opposite. The balls have the same mass, so the total momentum is $mv + m(-v) = 0$.

The balls collide and bounce apart with their velocities reversed so the ball originally moving at $+v$ is now moving at $-v$, and the ball originally moving at $-v$ is now moving at $+v$. I can draw a diagram if you want.

OK, yes

After the collision the total momentum is $m(-v) + mv = 0$. So the total momentum hasn't changed. Momentum is conserved. OK so far?

Yes

Now suppose that instead of standing still beside the table you are running at a velocity $v$ alongside one of the balls. In this frame the ball you are running alongside is motionless while the other ball is moving towards you at a speed $-2v$.

I suspect we might need a diagram for this ...

Ok

@yuvrajsingh The top diagram shows what the collision looks like when you are standing beside the table. The total momentum is zero and it is the same before and after the collision. OK so far?

Yes

Now suppose you are running alongside the blue ball at velocity $+v$. To get the velocities in your frame we have to subtract $v$ from all the velocities. So in your frame the blue ball is initially moving at $+v - v = 0$ and the red ball is moving at $-v - v = -2v$. The total momentum is $-2mv$. OK so far?

Yes

After the collision the red ball is stationary in your frame while the blue ball is moving at $-2v$. So the total momentum is $-2mv$. Same as before the collision. So in your moving frame the total momentum is conserved as well.

Yes, I got your point.

I need to go now