@JohnRennie: Hi sir. Good morning :-)

@GuruVishnu hi :-)

Sorry for the slow response. It's been a busy morning.

No problem sir :-)

@JohnRennie: Are you free now sir?

I'm having a doubt in the following question:

An accurate pendulum clock is mounted on ground floor of a high building. How much time will it lose or gain in one day if it is transferred to the top storey of the building which is 200 m higher than the ground floor?

@GuruVishnu It's just the variation of g with height.

@JohnRennie Yes sir. I understand that. I'm typing my doubt:

I understand due to g getting smaller and smaller, the time period decreases. Or a second for the clock takes longer and hence looses time.

But the problem is is a pendulum clock a seconds pendulum with time period of two seconds?

I looked at the internet, and it seems different pendulum clocks have different effective lengths which means different time periods.

As g gets smaller (i.e. as you go up) the period increases because $T = 2\pi\sqrt{\ell/g}$

Yes sir. That was how I arrived at that conclusion, i.e., the clock looses time.

If it's 4:30 it might show 4:00 or something like that.

The clock is designed to have n periods per day, for some value of n that we don't know, so when the period increases those n periods take longer and the clock runs slow.

Ok sir. So the value of n need not be the same as a seconds pendulum, right?

I just looked at the pictures of some pendulum clocks.

Correct. You just need the ratio $T_0/T_h$, where $T_0$ is the period at ground level and $T_h$ is the period at height $h$.

And that's just $\sqrt{g_h/g_0}$

For the sake of simplicity, I assumed T=1 s which in fact gave the correct answer: it looses 2.7 s for one day. Could you give a brief explanation on why we need the ratio $T_0/T_h$? Is that to find the loss in time per time period?

I rather took a long approach even with one second as the standard time period.

I used the difference per standard time period and multiplied with standard one day.

By standard, I meant, measuring time with say an atomic clock, or a digital watch.

Suppose the number of periods at ground level is n_0 and the period T_0, then we have $n_0 T_0 = 86400$ seconds (i.e. one day). OK so far?

Yes sir.

The clock is designed to count $n_0$ periods in one day, so if the period increases to $T_h$ it will now think one day lasts $n_0 T_h$ seconds. We need to find how many seconds that is.

And we know $T_h = T_0 \sqrt{g_0/g_h}$. Yes?

@JohnRennie Ok sir. It's even more clear to me now. We need to compute $n_0(T_h-T_0)$ right?

@JohnRennie Yes sir.

@GuruVishnu yes :-)

Ok sir :-)

I think it's not possible to find the time difference without the value of $n_0$.

So should the time period of a pendulum clock be one second or even if it isn't will that be totally up to the internal mechanisms which are not of our interest?

@GuruVishnu If we substitute for $T_h$ we get $n T_0(\sqrt{g_0/g_h} - 1)$. Yes?

Yes sir. And this gives how much excess time the clock reads compared to a standard clock.

Which is exactly the answer for the question.

And we known $n_0 T_0 = 86400$ seconds

So the excess time is just $\Delta t = 86400 \times (\sqrt{g_0/g_h} - 1)$

Apparently this is why assuming standard period as one second worked for me.

I was wondering why they didn't use a seconds pendulum instead.

Yes, the value of $T_0$ doesn't matter so you can assume any value and the calculation will work.

Ok sir. May I know how long will you be available?

Ok sir. I'd like to ask one last doubt.

OK

Question:

Two blocks of masses $m_1=1~\mathrm{kg}$ and $m_2=2~\mathrm{kg}$ are connected by a spring of spring constant $k=24~\mathrm{N/m}$ and placed on a frictionless horizontal surface. The block $m_1$ is imparted an initial velocity $v_0=12~\mathrm{cm/s}$ to the right. The amplitude of the oscillation is _____.

My approach:

Since there is no horizontal external force, the momentum of the system (spring+blocks) must be conserved. I observed the system from the centre of mass's reference frame. It initially travels with a speed of $4~\mathrm{cm/s}$ relative to ground and hence in it's frame, the block $m_1$ moves towards it with a speed of $8~\mathrm{cm/s}$ and $m_2$ moves towards right with the same speed as of COM i.e., $4~\mathrm{cm/s}$.

I also found the value of spring constants of the spring in between the blocks and the centre of mass.

Using energy conservation in this frame, I got the value of amplitudes of both blocks relative to the centre of mass. But none of them were equal to the options (1,2,3 and 4)

So, I wanted to ask how would we define the amplitude when the system doesn't oscillate in one frame due to it's horizontal motion?

(End of message)

Whenever you have a system like always consider working in the centre of mass frame. In this case it makes the question very easy.

Yes sir. I'd assume the centre of mass point as a fixed wall and sit on it and observe the system :-)

My doubt was, how do we define amplitude when both the blocks are oscillating and translating in the horizontal direction?

In the centre of mass frame the velocity of m1 is twice the velocity of m2, so in this frame m1 has velocity 8cm/s and m2 has velocity 4cm/sc

Calculate the total KE of both and equate this to the PE of the spring.

Yes sir. And this velocity relation ($v_1=2v_2$) is always true in order for the momentum to be conserved within the centre of mass frame.

I get $A_1=\frac{4}{3}$ cm and $A_2=\frac{2}{3}$ cm.

Does that match an answer?

The correct answer is $A_1+A_2=2$ cm. No problem. But when $m_1$ approaches minimum value, $m_2$ approaches maximum as their velocities are always in the same direction. So shouldn't this be the difference instead $A_1-A_2=\frac{2}{3}$?

Why do we need to consider the sum of amplitudes instead of the difference? The latter is the one we would be observing from the centre of mass frame, right sir?

Short version: Why are we concerned about $A_1+A_2$ instead of $|A_1-A_2|$?

@GuruVishnu we've been a bit cavalier about the signs of the velocities.

The velocities are always equal and opposite, so actually $v_1 = -2v_2$. Yes?

@JohnRennie Yes sir. I think that's where this discrepancy arises. Taking this into consideration, I hope the sum of amplitudes would make sense as both blocks either move towards each other or away from each other.

@GuruVishnu hi :-)

@JohnRennie Hi sir :-) Sorry for the late reply. I had to move away immediately to close the windows during that time. Antivirus (well I didn't find a better term for it) was sprayed in our street.

When I returned back, you had already left, so I reserved this once you arrive back.

I've seen videos of that. They spray a mist to sterilise all surfaces don't they?

You don't want it all over your computer :-)

Yes sir. It can't protect the computer against any malware :-)

:-)

Did you watch the entire live stream for the Crew Dragon launch sir?

I'm watching it in sections when I've have my breakfast/lunch/dinner.

It's interesting to see every bit of that event and listening to the details.

Do you have any ideas on this question of mine? : Why do Crew Dragon astronauts need to climb up one level using the stairs before ingress?

I have no idea I'm afraid.

I didn't watch the launch, but I'm hugely impressed that SpaceX have accomplished this.

Ok sir. No problem. Yes that's a great achievement!

Thank you for clearing that amplitude doubt of mine :-)

You're welcome :-)

Keep in mind that you should always consider using the COM frame in this sort of thing. In this case you can see that it simplifies the calcualtion so much that it becomes almost trivial.

Yes sir.