Recent content by brotherbobby

Thank you @Ibix. I should add, to help the admin, @Greg Bernhardt, that the formatting in ##\rm{\LaTeX}## does not disappear in the main text, where we type in our attempt and doubts, if we try to edit our message. Needless to say, this is also not a problem in all future replies. But for the...

Hi @pasmith, yes it seems I was mistaken too in my plotting on ##\verb|desmos.com|##. I took the constant ##k## with a positive ##(+)## sign. Of importance is the minus ''##-##'' sign before the term ##k## in the question, as you pointed out. If it was ##+k##, then the book is correct. As it...

Statement of the problem : Let me copy and paste the problem as it appears in the text. [Riley, K.F., Hobson,M.P., Bence,S.J. (2006) Mathematical Methods for Physics and Engineering. (3E)]. Attempt : One real root is guaranteed, as ##n =\;\text{odd}##. Differentiating ##f'(x) =...

Yes sorry about that. I agree. I am sorry I don't see how that is - the earth's motion, whether towards or away, affecting when the earth sees the moon disappear and reappear. I am sure it won't be easy for you to draw a diagram for it. Do you have a text with a diagram to which I can refer...

Let's have the image again. It might be obvious from the picture, but let me state it all the same. In 1 and 2, the object it moving away from Jupiter, while in 3 and 4, it is moving towards. In 1 and 3, the moon Io is seen by the earth to be sliding out of the eclipse, while in 2 and 4 it is...

Are you saying it's the distance between 3 and 4 that matters? According to the authors, it's not the distance between 1 and 2, or between 3 and 4. They could be as close as you wish. It's how the earth was moving at those points. Either towards or away from Jupiter.

No, see the diagram the authors have drawn. There is no position difference between the labelled points 1 (and 2) and that between 3 (and 4) far as distance from Jupiter goes. According to the authors' reasoning, at 1 and 2, the earth is moving away from Jupiter and at 3 and 4, it was moving...

To call it "inadequate" is putting it mildly. My calculation above (post #3) shows that if the authors had Doppler's effect in mind for the differing speeds of light when the earth was moving towards and away from Jupiter, they are wrong in a big way. It's not about how the earth is moving...

I think it would be fair on my part to carry out the calculation assuming the Doppler's effect, fair to the authors of the text that is. Let's assume the following was known at the time of R##\ddot{\text o}##mer. Data : The distance to Jupiter from earth, ##d_J = 5.58\times 10^8\;\text{km}##...

Statement of the problem : I copy and paste the clip from the text, underling in red the sentences where, in my opinion, the author(s) have gone wrong. I annonate to the right of the text in blue, assuming they are both readable. The text in question is titled Special and General Relativity...

Problem statement : I copy and paste the problem as it appears in the text, Conceptual Physics (12th Edition) by Hewitt, P. Two balls are released simultaneously from rest at the left end of equal-length tracks A and B as shown. Which ball reaches the end of its track first? Attempt : The...

Thank you. The point may be elementary but deep. Far as I know, books don't say these things. One has to "discover" them on one's own, or with help.

Yes, forgive me, I realised my mistake just after you posted. I was thinking of editing my response. Let me do it again. The velocity of the electron after 2 s is fine, ##v(2) = 0##. The position of the electron : ##\boxed{x(2)} = \dfrac{a}{2}+a-\dfrac{a}{2} = \boxed{a}##. The electron moves a...

Imagine the body to start from the origin. After 1 s, the body is at a position of ##x(1)=\dfrac{1}{2}a## with a velocity ##v(1) = a##. After another second of deceleration by the same amount, the position of the particle is ##x(2) = a-\dfrac{a}{2}= \dfrac{a}{2}##. As for i ts velocity : ##v(2)...

Yes, let me calculate. We have the position of the electron relative to the origin O : ##x(t) = \dfrac{a_0}{\omega}t-\dfrac{a_0}{\omega^2}\sin\omega t##. Upon differentiating, ##\dot x(t) = \dfrac{a_0}{\omega} - \dfrac{a_0}{\omega}\cos\omega t##. To check, this matches the initial condition of...