WAVES

At boundaries between different media, there is generally both reflection and transmission. A “boundary” is a place where there is a change in “how hard it is” for the wave to cause a disturbance the two media. This includes the end of a medium.

               • The incident wave is the one that approaches the boundary, but hasn’t reached it yet.

               • The reflected wave is the one that moves away from the boundary, but in the same medium as the incident wave.

               • The transmitted wave is the one that moves away from the boundary, on the other side of the boundary from the incident wave. The results depend on “how hard” it is to disturb the new medium compared to the old. More massive media are “harder to disturb.” Also, media with a stronger restoring force are “harder to disturb.”

The reflected and transmitted waves are described as inverted or upright and reversed or not.

              • Inverted means that, compared to the incident wave, the disturbance in the medium is the opposite. Transverse waves are turned upside down, compression turns into rarefaction, etc.

              • Upright means that the disturbance in the medium is the same as with the incident wave (NEITHER of these means “pointing up” or “pointing down”. They are relative to the incident wave.)

               • Reversed means turned around along the direction of travel.

             The following table shows the full continuum of possible cases. The first row described what the incident wave encounters as it reaches the boundary and looks at what is on the other side. (The shaded rows apply only in the common case where the restoring force is the same on both sides of the boundary.)

Reflection Of Waves :

(a)     Waves on reflection from a fixed end undergoes a phase change of 180°.

(b)     While a wave reflected from a free end is reflected without a change in phase.

1.         A pulse in a rope approaches a solid wall and it gets reflected from it

            The wave pulse after reflection is best represented by _____

Solution:

2.    A stretched wire will produce a note of very high frequency, then the nature of the wire is

Solution:

                               Thin and short wire of light material  under high tension

1.    The extension in a string, obeying Hooke's law is x. The speed of sound in the stretched string is V. If the extension in the string is increased to 1.5x the speed of the sound will be

Solution:

               \(V\alpha \sqrt{e}\,\) (Hooke’s law application)

            \(V=\sqrt{\frac{T}{M}}\,\propto \sqrt{\frac{YAe}{Kl}}\propto \sqrt{e}\)

                = 1.22 V

2.      A transverse wave propagating on a stretched string of linear density \(3\times {{10}^{-4}}kg{{m}^{-1}}\) is represented by the equation \(y=0.2\operatorname{Sin}(1.5x+60t)\) where x is in meter and t  is in seconds. The tension in the string in N.

Solution:

               \(n\lambda =\sqrt{\frac{T}{m}}\) after comparing with general equation

               \(y=a\sin \left( \omega t\,\,+\,\,kx \right)\)

1.      A uniform rope of  length 12m and mass 6 kg  hangs vertically from a rigid support . A block of mass 2kg is attached at the free end of the rope.  A transverse pulse of wavelength 0.06m is produced at the lower end of the rope.  The wavelength of the pulse when it reaches the top of the rope is

Solution:

               Tension at the bottom =2kg.Wt = \({{T}_{b}}\)

            Tension at the top =(2+6)=8 kgwt = \({{T}_{t}}\)

                                                     \(\frac{{{V}_{t}}}{{{V}_{b}}}=\frac{n{{\lambda }_{t}}}{n{{\lambda }_{b}}}={}{}\sqrt{\frac{{{T}_{t}}}{{{T}_{b}}}}\\\)

                                                     = 0.12 m

2.         A string of length l changes freely from a rigid support.  The time required by a transverse pulse to travel from bottom to half length of  the string is

Solution:

               \(t=2\sqrt{\frac{l/2}{g}}\)

                 = \(\sqrt{\frac{2l}{g}}\)

1.         A source emitting sound of frequency 180 Hz is placed in front of a wall at a distance of 2 m from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and the detector for which the detector detects a maximum of sound. Speed of sound in air = 360 m/s.

Solution:

               Path difference D = (SO + OD) – SD

The direct wave received by the detector from the source travels a distance SD and reflected wave travels a distance (SO – OD).

            \(\Delta =2\left[ {{\left\{ {{(2)}^{2}}+\frac{{{x}^{2}}}{4} \right\}}^{1/2}}-x \right]\ \text{metre}\)

For constructive interference D = l, 2l, …. …. The minimum distance will correspond to D = l

                   \(\lambda =\frac{v}{\nu }=\frac{360}{180}=2\ \text{metre}\)

                       \(2\left[ {{\left\{ {{(2)}^{2}}+\frac{{{x}^{2}}}{4} \right\}}^{1/2}}-x \right]=2\)

Solving, we get x = 3 meter.

2.      A string is stretched between fixed points separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

Solution:

               \(\frac{315}{420}=\frac{n}{n+1},\,\,n=3\)

               lowest frequency  \(=\frac{315}{3}=105Hz\)