Cool, thanks for your explanations guys. So we should say whether the acceleration is infinite or not has nothing to do with the amount of extra energy needed here, right?
I tried to take the analysis of using inelastic model further.
Suppose the mass of belt is M moving with speed v, and sand...
I can understand how it can be explained using the change of momentum to find the average force on the belt, then us $$P=Fv$$ to get the power.
What I didn't understand is why the energy solution won't work. If in 1 second, sand with mass m has fallen on the belt and accelerated to speed v...
This is an interesting idea. So do you mean that the sound we hear will automatically be "decomposed" into two sound waves with distinct frequencies? I think you are probably right, it is like our brain is embedded with a spectrum analyzer, isn't it?
Thank you. I think it makes sense. It is like a symmetry, isn't it?
For a fixed position x, we would obtain a similar graph for y changes with time t, so the frequency we would hear will be ##\frac{\Omega_1}{2\pi}##
Hmmm, brilliant, thanks
for a) I have ##v = \lambda f= \frac{\omega \lambda}{2 \pi}##
for c) and d) I denote ##\frac{2\pi}{\lambda_1} = k_1## and ##\frac{2\pi}{\lambda_2} = k_2## assuming ##k_1 > k_2##.
so using the triangular identity I got $$y_1 + y_2 = 2A cos(\frac{k_1+k_2}{2} x - \frac{\omega_1 + \omega_2}{2} t)...
Could it possibly be the condition that it is not doing steady circular motion (if we increase the angular frequency very slowly to its maximum), but instead, we give the ball a tangential impulse at first. So the ball will start doing circular motion with angular frequency ##\omega_i## but...
Yeah, you are right, thanks. I didn't notice R is actually extension here, so expression for R has to change. Besides, the graph given in question is also misguiding.
But this won't affect my answer b) and c) since I didn't use formula of extension R when I derive answer in b) and R never gets...
Thank you my first replay here in PF:)
Yes, but what I can see is only that the denominator cannot be 0. i.e. $$k-m\omega^2≠0$$ so you are suggestion the limit is ##\omega_i^2=\frac{k}{m}##?
But there is always $$\omega_c<\omega_i=\sqrt{\frac{k}{m}}$$ and the question requires us to find a limit...
and this is my solution
for question (d), it may seems that $$R=(k)/(k-m\omega^2)R_0$$ so that $$\omega ≠ \omega_i =√(k/m)$$
but $$\omega_c <\sqrt{k/m}$$ is always true, ##\omega_i## corresponds to the limit case when ##F_max## is infinitely large
Besides, I don't know other Physics prevents...