Recent content by Ark236

Thank you very much :D.

Thanks, when I use $y_i = A_i sin(\omega t - k_1 x )$ and $y_r = A_r \sin(\omega t) + k_1x)$ the calculations work. However, it is not clear to me why it does not work in the other case.

thanks, I made a mistake when copying the calculation. y_r = (mu2-mu1)/(mu1+mu2)y_i and y_t = A_t sin(k2 x - wt). However, if mu2>mu1 i expect that y_r will negative since the reflected wave is inverted.

To obtain the amplitude of the reflected and transmitted wave, I consider that the initial pulse is traveling to the right and I use the boundaries condition: 1. y_i(x = 0) + y_r(x = 0) = y_r(x = 0) 2. dy_i/dx |_(x = 0) + dy_r/dx |_(x = 0) = dy_r/dx |_(x = 0) The expression for the incident...

I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards. thanks image was obtained from here Thanks.

Thank you very much for the help

Thanks, I think it's a little clearer now. The problem is that the force that body A exerts on body B is not m_A g since the system is accelerating. However, for me is still not easy to see to assume that, because similar to the elevator problem the person is exercising the real weight over...

Hi, Regarding TSny's answer. For me is correct, the force of body A that acts on body B is m_A g. In the Body B there is one normal, N_b, that is the reaction of the surface between the body B and the plane. Regarding Kuruman's answer, if I do it that way I immediately get the correct result...

In order to calculate the acceleration, I sum the equations sum fx^B and sum fx^A. F_{r,B} - m_Ag sin(theta) - (m_A+m_B) g sin(theta) = - (m_A+m_B) a then a = -[F_{r,B} - (2 m_A+m_B) g sin(theta)]/(m_A+m_B) =-[muk (m_A+m_B) g cos(theta) - (2 m_A+m_B) g sin(theta)]/(m_A+m_B) but the...

Thanks so much for the answer. Now for me is clear :D .

Because from the definition, the integration limits are the initial and the final points. From path c the particle starts from the point (0,1) and ends at the point(1,0). Also, the ##\vec{dr}## is the direction of the displacement, in that case, would be negative (##\vec{dr} = -dx \hat{i}##).

So how would you set up the integral of this path?, $$ W = \int_{1}^{0} \vec{F}\cdot d\vec{r} =\int_{1}^{0} (xy \hat{i} +y^2\hat{j })\cdot (dx \hat{i}) = \int_{1}^{0} xy dx $$ For me it is not totally clear yet the reason.

Thanks for the answers. It is clear to me that the integral must be negative since the force is opposite to the displacement. However, the path integral is defined as $$ W = \int_{a}^{b} \vec{F}\cdot d\vec{r} $$ where a and b refer to the start and end points, respectively. In this example a...

In the book it is mentioned that, in path c, the line integral would be: $$\int \vec{F}\cdot \vec{dr} = A \int_{1}^{0}xy dx = A\int_1^0 x dx = -\dfrac{A}{2}$$. but I think that dx is negative in that case, the result would be positive, right?

I study of interaction between a system with a reservoir considering a weak coupling between them. I consider a bosonic bath, the initial state are separable and the operator of interaction between the system and bath is linear in the displacements of the oscillators. . In the book "Quantum...