Reflection and Transmission of String Waves

[Ark236]

Homework Statement

Suppose that two wire have different densities and are joined at x = 0. Determine the amplitude of the transmitted and reflected wave at x = 0

Relevant Equations

y_i = A_i sin(k1 x - wt)
y_r = A_r sin(k1 x +wt)
y_t = A_r sin(k2 x - wt)

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, reflected and transmitted wave are:

y_i = A_i sin(k1x - wt) wave traveling to the right.
y_r = A_r sin(k1x + wt) wave traveling to the left.
y_t = A_r sin(k2x - wt) wave traveling to the right.


The first equation lead to y_i - y_r = y_t and the second equation lead to y_i k1 + y_r k1 = y_t k2. When I solve the system of equation I find that the amplitude of the reflected wave is:

y_r = (k2-k1)/(k1+k2)

which can be express in terms of the linear mass density mu of each medium using that k2 = k1 sqrt(mu2/mu1):

y_r = (mu2-mu1)/(mu1+mu2)

This result is wrong because if mu2>mu1 the reflected wave would be inverted with respect to the incident. There is a minus sign in some place.

thanks

 

[Steve4Physics]

Homework Statement: Suppose that two wire have different densities and are joined at x = 0. Determine the amplitude of the transmitted and reflected wave at x = 0
Relevant Equations: y_i = A_i sin(k1 x - wt)
y_r = A_r sin(k1 x +wt)
y_t = A_r sin(k2 x - wt)

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, reflected and transmitted wave are:

y_i = A_i sin(k1x - wt) wave traveling to the right.
y_r = A_r sin(k1x + wt) wave traveling to the left.
y_t = A_r sin(k2x - wt) wave traveling to the right.


The first equation lead to y_i - y_r = y_t and the second equation lead to y_i k1 + y_r k1 = y_t k2. When I solve the system of equation I find that the amplitude of the reflected wave is:

y_r = (k2-k1)/(k1+k2)

which can be express in terms of the linear mass density mu of each medium using that k2 = k1 sqrt(mu2/mu1):

y_r = (mu2-mu1)/(mu1+mu2)

This result is wrong because if mu2>mu1 the reflected wave would be inverted with respect to the incident. There is a minus sign in some place.

thanks

If A_t is the amplitude of the transmitted wave then
y_t = A_r sin(k2x - wt) should be
y_t = A_t sin(k2x - wt).

Edit. Also "y_r = (mu2-mu1)/(mu1+mu2)" can not be correct as it is inhomogeneous. The left side has the dimension of length but the right side is dimensionless.

 

[Ark236]

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.

 

[TSny]

You wrote the incident and reflected waves as $$y_i = A_i \sin(k_1x - \omega t)$$ $$y_r = A_r \sin(k_1x+ \omega t)$$ Consider $x = 0$ for these two equations. If $A_i$ and $A_r$ are both positive, is the reflected wave inverted relative to the incident wave or not inverted?

Some books write the waves as $y = A \sin(wt \pm kx)$ with the $\omega t$ coming first in the argument of the sine function. In this case consider the incident and reflected waves written as $$y_i = A_i \sin(\omega t -k_1x )$$ $$y_r = A_r \sin(\omega t+k_1x )$$ Now is the reflected wave inverted if both $A_i$ and $A_r$ are positive?

 

[Steve4Physics]

I deleted my reply as I decided it was confusing.

 

[TSny]

1. y_i(x = 0) + y_r(x = 0) = y_t(x = 0)
2. dy_i/dx |_(x = 0) + dy_r/dx |_(x = 0) = dy_t/dx |_(x = 0)

These equations are to hold at all instants of time.

The first equation lead to y_i - y_r = y_t and the second equation lead to y_i k1 + y_r k1 = y_t k2.

You should be looking for equations relating $A_i$, $A_r$, and $A_t$. You need to solve for $A_r$ and $A_t$ in terms of $A_i$.

 

[Ark236]

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.

 

[TSny]

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.

In the case where you are using $y = A\sin(kx \pm\omega t)$, show that the reflected wave is inverted if $A_r$ and $A_i$ have the same sign. See the first paragraph of post #4.

If you use $y = A\sin(\omega t \pm kx)$, show that the reflected wave is inverted if $A_r$ and $A_i$ have opposite sign.

It might help to note that $\sin(\omega t - kx) = -\sin(kx - \omega t)$, but $\sin(\omega t + kx) = +\sin(kx + \omega t)$

 

[Ark236]

Thank you very much :D.