Wave Reflection and Stationary Waves

$Y = 10\sin \dfrac{{\pi x}}{4}\cos \pi t$

At a free and, the wave is reflected as it is with just as change in its direction of propagation 
$y= A  \sin  ( \omega t+kx)$
$y _r= A  \sin  (\omega t-kx)$
phase difference in time domain $= 0$

Regarding open pipe both ends are pressure nodes.

An antinode is a point on the string, where the amplitude at that point is maximum 

The correct option is (a)

In a free boundary, the restoring force is always zero

The correct option is (a)

In reflection of a wave at open boundaries, the phase reversal dosent take place

The correct option is (c)

Given frequency $\nu$=100 H. 

Since the wave is travelling from low to more dense medium, thus there will be inversion of the reflected wave . Since it travels back from origin, thus $40x$ will become $-40x.$ Also, amplitude of reflected wave is given by:
${ A } _{ r }=\dfrac { Z _{ 1 }-{ Z } _{ 2 } }{ { Z } _{ 1 }+Z _{ 2 } } A.\quad Z=\mu c=\mu \sqrt { \dfrac { T }{ \mu  }  } =\sqrt { \mu T } \ Thus\quad { A } _{ r }=\dfrac { \sqrt { \mu T } -\sqrt { 4\mu T }  }{ \sqrt { \mu T } +\sqrt { 4\mu T }  } A=\dfrac { -1 }{ 3 } A\ $
Thus it become $-6/3=-2$.

Statement (i) is incorrect because the resultant wave's amplitude changes, though frequency/wavelength remain the same. Statement (iii) is incorrect because mechanical and electromagnetic waves can superpose (e.g., light interacting with a medium). Statement (ii) is incorrect as the resultant amplitude depends on phase. Statement (iv) is incorrect as superposition occurs for waves of different properties. Given the options, (i) and (iii) are the most clearly incorrect statements regarding the fundamental principles of superposition.

There are three things we need to take into account:

• Energy transfer
• Change in velocity
• Change in phase

We know that power delivered is proportional to $A^{2}$
Hence if power(energy) reduces to 64%. We get that Amplitude must reduce to 80% or 0.8A.
Now the reflected wave is moving in the opposite direction. (velocity is negative now).
Also because of the hard soft boundary reflection (there is a phase lag of $180^{\circ}$
Hence the new equation becomes:
$y = 0.8A sin(kx + \omega t + 30^{\circ} + 180^{\circ})$
Hence option B.

A pulse of wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of $\pi$ with the incident wave but wave velocity after reflection reverses its direction.

Since wave length is $2m$, half of wavelength is $1m$. Node forms after each half of wave length. So, node will be formed at each $1 m$
So, as node is formed at $3 m$
next node will be formed at $3+1=4\ m.$

$v=f\lambda$
$\lambda=\dfrac{340}{1360}=\dfrac{1}{4}=0.25m$
$=25cm$
As the end is a node, shortest distance at which antinode is formed is $\dfrac{\lambda}{4}=6.25cm$