Recent content by Saptarshi Sarkar

I know that if the transformation was canonical, the form of Hamilton's equation would remain invariant. If the generating function for the transformation was time independent, then the Hamiltonian would be invariant and we could directly replace q and p with the transformation equations to get...

I was thinking that the angular momentum must be independent of time as it looks like a trajectory under a central force. Is it not true? For the other part, I did a mistake. Thanks for pointing it out. I need to consider the tangential velocity as well. Total linear momentum =...

I know that the force must be a central force and that under central forces, angular momentum is conserved. But I am unable to mathematically show if the angular and linear momentum are constants. Radial Momentum ##p=m\dot r = ma\dot \theta=ma\omega## Angular Momentum ##L=mr^2\dot\theta =...

I get that. But I don't understand why the element is equal to the wavelength and not half the wavelength. From the diagram of a standing wave, the distance between two antinodes is equal to half the wavelength.

I was reading a pdf on acoustic grating for my practicals when I saw that the grating element of an acoustic grating is equal to the wavelength of the sound wave. I also checked a few other sources and got the same. I do not understand why. I know that for an acoustic wave, the standing wave...

Attempt: By drawing the Free Body diagrams and calculating the different tensions, I got the following results ##T_1=\frac{(M_1+M_2)}{2}g## ##T_2=\frac{\sqrt 3(M_1+M_2)}{2}g## ##T_3=M_2g## But, I am not sure what the answer is as although ##T_2>T_1## but ##T_3## does not depend on ##M_1##...

I first tried to get the solution by conserving the rotational kinetic energy and got ##\omega'=\frac2{\sqrt5} \omega##. But, it was not the correct answer. Next I tried by conserving the angular momentum and got ##\omega'=\frac 45 \omega##, which is the correct answer. Why is the rotational...

My Attempt :We need to maximize ## D=\sqrt{x^2+(y+2)^2} ## subject to the constraint ##4x^2 + 5y^2 = 20##. From the constraint equation, we can write ##x^2=\frac{20-5y^2}{4}## Using this in the formula for distance, ##D=\sqrt{\frac{20-5y^2}{4}+(y+2)^2}## Differentiating this wrt y, and...

I read about the tunnel theory of alpha decay in the book by Arthur Beiser. Here are a few lines from the book. "Here is a plot of the potential energy U of an alpha particle as a function of its distance r from the center of a certain heavy nucleus. The height of the potential barrier is...

I am having trouble understanding why any positivity charged particle like the proton or the alpha particle should have a coulomb barrier which prevents it from leaving directly. Only the nuclear force can do that . There is no negetive charges inside the nucleus that is trying to keep it inside...

Shouldn't the barrier be less deep due to the Coulomb repulsion between the protons in the nucleus? Why then should it be raised above? I am unable to understand why the proton needs higher energy to leave the nucleus than the neutron. They both are under the same nuclear attractive force but...

I was reading an introductory text on nuclear models and came across the Fermi Gas model. I understand that the depth of the potential well of the proton should be less than the depth of the potential well of the neutron due to the Coulombic repulsion between the protons. But I did not...

Thanks !

The question :- My attempt :- The confusion that I am having is that to get the required form of the equation of motion, I had to approximate ##\theta## to be small to get ##x=l\theta## so that I could get the acceleration and the velocity. But, I had to leave the ##sin(\theta)## in the...

I know that the potential of a simple pendulum is given by the above formula and that we can expand ##cos\theta## to get ##V=mgl\left(\frac{\theta^2}{2}-\frac{\theta^4}{24}+...\right )## I am guessing that the answer is ##\theta^4##, but I am not sure what "order" means here.