# Reflection of Waves and Formation of Standing Waves

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1. Nov 24, 2015

### FreezingFire

I have three doubts in regard to waves on a string which I will try to make as clear as possible.
For this purpose, I have considered a general wave:
$$y_i=y_0\sin(\omega t - kx)$$

(1) If a wave pulse:
$$y = y_0 e^{\frac{-1}{T^2} \left(t-\frac xv \right)^2}$$
is incident against a rigid boundary, such as a fixed wall to which the string is attached, why does it get inverted when it is reflected? I know that as the wave hits the boundary, considering the wave pulse travelling from left to right, the left part pulls the element of string fixed to the wall upwards, and the wall exerts an equal and opposite force on it, thus preventing the element from moving. Due to this force, the left part of string is pulled downwards upto its mean position, beyond which it (according to me) moves downwards due to inertia. My question is, at this stage, won't the left part of the string pull the said element downward again? Wouldn't the wall again generate a new pulse in the upward direction? Since this doesn't really happen, what is the correct reasoning behind the inversion of the pulse? Also if possible, could similar reasoning be applied for reflection against a non-rigid (soft) boundary? (View attached image)

(2) Now considering the initial sine wave, if it reflects against the rigid boundary, what will be the equation of the reflected wave? As far as I know, it must be:
$$y_r=y_0\sin(\omega t + kx + \pi)$$
or,
$$y_r=-y_0\sin(\omega t + kx)$$
Is this correct? And what about reflection against a non-rigid (soft) boundary? Is it the following?
$$y_r=y_0\sin(\omega t + kx)$$

Also, how do we obtain these equations (short proofs or conceptual proof)?

(3) In our textbook, the standing waves were explained as superposition of two waves travelling in the opposite direction, where one was the incident wave (given as $y_i=y_0\sin(kx - \omega t)$ in the book), and the other was reflected from either a hard boundary or a soft one. But in both cases, the equation of standing wave used is exactly the same, i.e.:
$$y=2y_0\sin(kx)\cos(\omega t)$$
where component waves were given as $y_i=y_0\sin(kx - \omega t)$ and $y_r=y_0\sin(kx + \omega t)$. This meant that my equation for reflection was wrong too, as $y_r$ is different in this case! Also for soft boundary the exact same equation of standing wave was used, meaning the reflected wave was also the same in both cases! Clearly this isn't possible. So what is the actual mathematical derivation of standing waves on string fixed at both ends, and string fixed at one end only?

Finally, can we apply all these results for a longitudinal pressure waves (i.e. sound)?

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2. Nov 24, 2015

### .Scott

Maybe I can explain it this way.

You have a cord where a pulse is generated at one end "A" and then travels to the other end "B" where the cord is solid fixed to a wall.

But let's look at a similar situation: You have a cord where a pulse is generated at one end "A" and then travels to the midpoint "B" where it meets another inverted pulse traveling from the opposite direction an originating at "C", the other endpoint. In such a case, the pulse traveling from A to C will pass right through the pulse travelling from C to A - each escaping unchanged once the pass the center point B and each other.

But what would they look like at B? How would point "B" move? The answer is, it wouldn't. The two waves cancel each other out at that point. It is exactly as if "B" was fixed to a stationary wall.

3. Nov 24, 2015

### FreezingFire

Thanks for the detailed answer, but I already knew this explanation from our textbook! What I further need is how do we explain this in terms of forces only [this is for doubt (1)]. I now realise that maybe this delves deep into how a wave actually propagates on a string in terms of forces....!

Last edited: Nov 24, 2015
4. Nov 25, 2015

### FreezingFire

Anyone there? Please help me with all my three doubts!

5. Nov 25, 2015

### FreezingFire

Also, I see that perhaps my doubt #3 is a bit confusing. So I will reword it:

1. Standing waves on string fixed at both ends
Consider a wave $y_i = y_0 \sin(kx - \omega t)$ on a string fixed at $x=0$ and $x=L$.

As the incident wave gets reflected from the fixed end at $x=L$, the equation of reflected wave is $y_r = y_0 \sin(kx + \omega t)$, as the reflected wave in this case travels in the negative x direction and has a phase difference of $\pi$ with the original wave. These two waves interfere to form the standing wave:
$$y = 2y_0 \sin(kx) \cos(\omega t)$$
This is correct as far as my textbook is concerned.

2. Standing waves on string fixed at one end
Again consider the same wave $y_i = y_0 \sin(kx - \omega t)$ on a string fixed at $x=0$ and attached to a light string (acting as a free end) at $x=L$.

Now, the incident wave reflects at $x=L$ as $y_r = -y_0 \sin(kx + \omega t)$, as it has no phase difference with respect to the original wave, and it travels in the negative x-direction. These two waves would interfere to give the standing wave:
$$y = -2y_0 \cos(kx) \sin(\omega t)$$
This would imply that $x=0$ is an antinode, which is clearly impossible. So where am i going wrong? I think the error might be in the italicized sentence. Also, our textbook has directly given us the equation of standing wave in this case also as:
$$y = 2y_0 \sin(kx) \cos(\omega t)$$
which is same as for a string fixed at both ends.
This is why i was actually asking for the equations and derivations of reflected waves in doubt #2.

6. Nov 25, 2015

### FreezingFire

Anyone there?

7. Nov 27, 2015