Wave Motion and String Waves Concept Page - 11

Example

Infer and use variation of amplitude with radius for a wave originating from a point source

Example:
A point source emits a sound equally in all directions in a non-absorbing medium. Two points P$P$ and Q$Q$ are at a distance of 9$9$meter and 25$25$ meter, respectively from the source. Find the ratio of amplitudes of the wave at P$P$ and Q$Q$.

Solution:
The intensity I$I$ of a sound wave at a distance r$r$ from a point wave source is given by ,I∝1/r2$I\propto 1/r^2$ ,
and intensity of a wave is related with amplitude by,                        
I∝A2$I\propto A^2$ ,if we combine these two relations , we get ,                       
A∝1/r$A\propto 1/r$
let AP$A_P$ and AQ$A_Q$ are the amplitudes of waves at point P and Q ,
therefore ,  AP/AQ=rQ/rP$A_P/A_Q=r_Q/r_P$ ,
given  rP=9m,rQ=25m$r_P=9m,r_Q=25m$ ,
therefore            AP/AQ=25/9$A_P/A_Q=25/9$ ,
or                      AP:AQ=25:9$A_P:A_Q=25:9$

Example

Example of wave equation for a point source

Example: A person sitting at a large distance from a point source of sound writes
down the following equation for the waves received by him y=Asin(2πx+700πt)${\displaystyle y=A\sin (2\pi x+700\pi t)}$ where x$x$ is in metre and t in sec. If the Bulk modulus of air  =1.40×105${\displaystyle =1.40\times {10}^5}$ Nm−2${\displaystyle N{m}^{-2}}$ then find the density of the medium.

Solution:
Since y=Asin(2πx+700πt)=Asin(Kx+ωt)${\displaystyle y=A\sin (2\pi x+700\pi t)=A\sin (Kx+\omega t)}$
∴k=2π,ω=700π${\displaystyle \therefore k=2\pi ,\omega =700\pi }$
∴${\displaystyle \therefore }$ velocity of sound v=ω/k=350m/s${\displaystyle v=\omega /k=350m/s}$
Givne that B=1.4×105${\displaystyle B=1.4\times {10}^5}$
∴v=Bρ−−√${\displaystyle \therefore v=\sqrt{\frac{B}{\rho }}}$
ρ=B/v2=1.40×105(350)2=87kg/m3${\displaystyle \rho =B/v^2=\frac{1.40\times {10}^5}{(350{)}^2}=\frac{8}{7}kg/m^3}$

Example

Intensity at a certain distance from a line source

Example: Find the relation between the intensity (I$I$) of a wave and on the distance (r$r$) from a line source.

Solution:
For a line source intensity at a certain distance r$r$ is given by:
I=pA=p(2πrl)$I={\displaystyle \frac{p}{A}}={\displaystyle \frac{p}{(2\pi rl)}}$
∴I∝1r$\therefore I\propto {\displaystyle \frac{1}{r}}$
where p$p$ is the power of the line source,
l$l$ is the length of the source
and r$r$ is the distance at which the intensity is found.

Example

Reflection of sound

Sound waves (like any other wave) return back in the same medium obeying the laws of reflection. The only requirement for the reflection of the sound wave is that the reflecting surface must be bigger than the wavelength of the wave. Reflection of sound wave is seen in echoes, megaphone, ear trumpet etc.

Definition

Incident, Reflected and Transmitted waves at a boundary

If a pulse is introduced at the left end of the rope, it will travel through the rope towards the right end of the medium. This pulse is called the incident pulse since it is incident towards (i.e., approaching) the boundary with the pole. When the incident pulse reaches the boundary, two things occur:
1. A portion of the energy carried by the pulse is reflected and returns towards the left end of the rope. The disturbance that returns to the left after bouncing off the pole is known as the reflected pulse.

2. A portion of the energy carried by the pulse is transmitted to the pole, causing the pole to vibrate.

Example

Write the equation of reflected wave for a given incident wave at a rigid boundary

Example: Find the phase change between incident and reflected sound wave from a fixed wall by writing equations of incident and reflected wave.

Solution:
When a wave is reflected from a wall, the wave is inverted . As the phase difference between a wave and its inverted wave is n¯$\overline{n}$, the phase difference is π$\pi$
y=A cos(kx+ωt)$y=Acos(kx+\omega t)$
yr=−A cos(kx+ωt)$y_r=-Acos(kx+\omega t)$
=A cos(kx+ωt+π)$=Acos(kx+\omega t+\pi )$
⇒ϕ=π$\Rightarrow \varphi =\pi$

Example

Incident, reflected and transmitted sound waves at a boundary

The simplest situation of reflection and transmission occurs when waves are impinging normal to the surface. The case of a longitudinal wave incident on the interface between two media is shown in the above figure.

Result

Equation of reflected wave for a given incident sound wave

Example: The displacement of the medium in a sound wave is given by the equation; y1=Acos(ax+bt)$y_1=Acos(ax+bt)$ where a & b are positive constants.
The wave is reflected by an obstacle situated at x = 0 . The intensity of the reflected wave is 0.64 times that if the incident wave. The equation for the reflected wave is given as y2=±s10Acos(ax−bt)$y_2=\pm {\displaystyle \frac{s}{10}}Acos(ax-bt)$ Find s$s$

Solution:
equation for sound intensity is I=12ρω2A2v$I={\displaystyle \frac{1}{2}}\rho {\omega }^2{A}^{2}v$
Ir=0.64Ii$I_r=0.64I_i$
12ρω2Ar2v=0.642ρω2Ai2v${\displaystyle \frac{1}{2}}\rho {\omega }^2{A_r}^{2}v={\displaystyle \frac{0.64}{2}}\rho {\omega }^2{A_i}^{2}v$
Ar2=0.64Ai2${A_r}^2=0.64{A_i}^2$
Ar=0.8Ai=0.8A$A_r=0.8A_i=0.8A$
s10A=0.8A${\displaystyle \frac{s}{10}}A=0.8A$
s=8$s=8$

Example

Reflection of a given incident sound wave in an open tube

Example: A train of sound waves is propagated along a wide pipe and it is
reflected from an open end. If the amplitude of the waves is 0.002 cm, the frequency 1000 Hz and the wavelength 40 cm, Find the amplitude of vibration at a point 10 cm from open end inside the pipe.

Solution:

Wavelength is 40cm
Amplitude to be found at 10 cm =404cm=λ4$={\displaystyle \frac{40}{4}}cm={\displaystyle \frac{\lambda }{4}}$
In open pipe from open pipe at a distance of λ/4${}^{\lambda }{/}_4$ always a node is formed. So, amplitude at 10 cm from open end will be zero because amplitude at node is zero.

Law

Laws of reflection of sound

Law 1 : Incident wave,normal to the reflecting surface and reflected wave at the point of incidence all lie in the same plane.
Law 2: Angle of incidence is equal to the angle of reflection.

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