Discussion on SHM

Jasmine

Sep 17, 2019 08:17

There are two particles doing SHM having same A but w as w1 , w2 we need to find time when they meet and initial phase is 0

Our Sir told us the solution with relative angular frequency (I dont get how )

John Rennie

So we have two equations of motion:
$x_1(t) = A\sin(\omega_1 t)$
$x_2(t) = A\sin(\omega_2 t)$

At time zero $x_1(0) = x_2(0) = 0$

And we want to find the next time when $x_1 = x_2$. Yes?

Jasmine

@JohnRennie $ sin (w_2t)=sin(w_1t)$

John Rennie

@Jasmine yes

Let's suppose $\omega_2 > \omega_1$

Jasmine

That gives $w_2t=nπ+{(-1)}^nw_1t$

John Rennie

@Jasmine Yes

So far it looks as if you're doing fine with this ...

yuvraj singh

Sep 17, 2019 08:24

Sorry for interr

Jasmine

@JohnRennie so the first time they meet is $t=\dfrac{π}{w_2+w_1}$

But the answer given is ${\dfrac{2nπ}{w_2-w_1}}$

yuvraj singh

Question become easier if we think about relative angular frequency and assume on e particle to be at rest

John Rennie

@Jasmine When the two particles meet they can be going in the same direction or opposite directions.

Those two cases are why you have the $(-1)^n$ in your equation.

When $n$ is even they meet going in the same durection and when $n$ is odd they meet going in opposite directions.

Jasmine

@yuvrajsingh yup this is what Sir said I dont no how can we apply concept of relativity to scalars

John Rennie

The answer you've been given tells you when they meet going in the same direction.

Jasmine

Sep 17, 2019 08:28

@JohnRennie Ok !

John Rennie

@Jasmine I can explain it with a diagram if you want.

Jasmine

But no information as such is mentioned

@JohnRennie Yes

yuvraj singh

@Jasmine done T=2pi/omega and we assume that w1 is greater than omega 2 then we take particle 2 at rest, let now our particle 1 moves with different angular velocity time period for complete revolution is 2pi/w1-w2 yes.

John Rennie

@Jasmine OK, give me a few moments to draw the diagram ...

Jasmine

@JohnRennie Ok

@JohnRennie why the even one for same direction can you explain

@yuvrajsingh angular frequency is scalar right

yuvraj singh

Sep 17, 2019 08:32

@Jasmine yes

Jasmine

@yuvrajsingh how can you apply relativity to them ! Or can you

yuvraj singh

@JohnRennie Iangualr frequency is scaler but angular velocity is not what I have written is angular velocity not frequency

John Rennie

yuvraj singh

T is time for one revolution w relative and 2pi is distance

John Rennie

@Jasmine you've probably seen this before, but SHM is related to circular motion.

Jasmine

Sep 17, 2019 08:36

@JohnRennie I can see the, image is it Phasor

John Rennie

If we have a particle moving around a circle at constant angular velocity $\omega$ then the position on the $y$ axis is $y(t) = A\sin(\omega t)$ so that gives us the SHM. Yes?

@Jasmine you can't see the image?

yuvraj singh

@JohnRennie can you check my answer I assume projection of the shm circular motion is it worng

Jasmine

@JohnRennie I could see when I clicked on it :-)

John Rennie

@Jasmine OK :-)

Jasmine

@JohnRennie yes

John Rennie

Sep 17, 2019 08:38

So now let's draw the two oscillators on the circle (I'll need a moment to update the diagram)

Jasmine

@yuvrajsingh Ohh now I understand you applied relative angular velocity in circular motion

@JohnRennie Ok

yuvraj singh

Yup

@Jasmine If I am not wrong you are unaware about projection of shm as circular motion.

John Rennie

@Jasmine yes, @Yuvraj and I are saying the same thing.

I'm just illustrating it graphically.

Jasmine

@yuvrajsingh nope I am aware but I like trigonometry more

yuvraj singh

@Jasmine hi are good in chemistry

Jasmine

Sep 17, 2019 08:45

23 mins ago, by Jasmine

That gives $w_2t=nπ+{(-1)}^nw_1t$

Suppose I get this equation so now when n is even they are going in different directions

yuvraj singh

@Jasmine I have issue in chemistry , I am going to kvpy this year will help me in some

Jasmine

@yuvrajsingh I can try

yuvraj singh

@Jasmine right now is my class, can I talk to you later around 7 of clock

Jasmine

@JohnRennie How did you know that even n gives the value when they are going in the same direction without phasor

@yuvrajsingh Ok there is one chemistry room as well you can post there

John Rennie

@Jasmine If you look at my diagram the first occasion when the amplitude of the particles looks like this (diagram incoming):

Jasmine

Sep 17, 2019 08:54

@JohnRennie that means we have to use phasor

@yuvrajsingh I will try but please ping me then

John Rennie

@Jasmine there

At this point the red particle is still going up the y axis while the blue particle is coming down the y axis. Yes?

Jasmine

@JohnRennie yes

John Rennie

@Jasmine I haven't labelled the angles because the diagram was getting messy, but if you look at the two angles $\omega_1 t$ and $\omega_2 t$ you'll see that $\omega_1 t + \omega_2 t = \pi$

That was the result you got:

$$ t = \frac{\pi}{\omega_1 + \omega_2} $$

yuvraj singh

@JohnRennie hi

John Rennie

@yuvrajsingh hi

yuvraj singh

Sep 17, 2019 09:04

Can I ask doubt about angular spectrum

John Rennie

@yuvrajsingh yes

yuvraj singh

Is it same as electromagnetic spectrum

Jasmine

@JohnRennie Ok

John Rennie

@Jasmine then the second occasion they meet is when the blue particle has gone all the way round the circle and caught up with the red particle i.e. $\omega_2 t = \omega_1 t + 2\pi$.

This time the particles are moving in the same direction on the $y$ axis.

That gives you the time:

$$ t = \frac{2\pi}{\omega_2 - \omega_1} $$

Which is what the answer you cited gave.

Jasmine

@JohnRennie ok so still its not good to generalise with 2nπ thingy

Because that means loss of many times they will meet

John Rennie

Sep 17, 2019 09:10

@Jasmine the $2\pi n$ equation is fine, but it gives you only the meetings when the particles are moving in the same direction i.e. when $\omega_2 t = \omega_1 t + 2\pi n$.

@Jasmine yes, you miss half the meetings.

Jasmine

Thank you @JohnRennie I got it :-)

John Rennie

I must admit it is mangling my mind slightly trying to visualise the opposite direction meetings ...

The first one is obvious from my diagram, but the second, third, etc are less obvious.

Really we need an animated diagram :-)

Jasmine

@JohnRennie thats why I dont like phasors

While using trigonometry is straightforward as we need to keep changing n and then get values of time

John Rennie

I think both approaches have their uses ...

yuvraj singh

Hi @JohnRennie

John Rennie

Sep 17, 2019 09:19

@yuvrajsingh hi

Jasmine

@JohnRennie if they had the same w and different initial phase then that will that mean they will meet only when they are going in different directions

yuvraj singh

Let jasmine finish then sir you ping me I will wait @JohnRennie

John Rennie

@Jasmine correct!

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