Wave Motion and String Waves Concept Page - 10

Definition

Define and find fundamental mode or first harmonic for standing wave on a string fixed at both ends

Fundamental mode/ First Harmonic : The lowest possible frequency at which a string could vibrate to form a standing wave pattern is known as the fundamental
frequency or the first harmonic.

Example: The string of a Sonometer is plucked so as to make it to vibrate in one segment. What is the frequency produced?

Solution:
l=(2n+1)λ2$l=(2n+1){\displaystyle \frac{\lambda }{2}}$ for a string fixed at one end.
When a string is made to vibrate in one segment, then length l=λ2$l={\displaystyle \frac{\lambda }{2}}$
ν=v4L  =νo$\nu ={\displaystyle \frac{v}{4L}}={\nu }_o$
And, (2n+1)=1$(2n+1)=1$   ⟹$⟹$ n=0$n=0$
n=0$n=0$ corresponds to first harmonic.

Definition

Define Higher harmonics and overtones for standing wave in a string fixed at both ends

The natural frequencies for the normal modes of oscillations of the system is:
ν=nv2L$\nu ={\displaystyle \frac{nv}{2L}}$

Putting n=1$n=1$ gives fundamental mode of vibration which is first harmonic.
n=2,3,4..$n=2,3,\mathrm{4..}$ gives higher harmonics. (Technically, 2nd, 3rd harmonic.. or 1st, 2nd overtone.... respectively)

Definition

Define and find fundamental mode or first harmonic for standing wave on a string fixed at one end

Standing waves can be produced on a string which is fixed at one end and whose other end is free to move in a transverse direction. Such a free end can be nearly achieved by connecting the string to a very light thread. If the vibrations are produced by a source of correct frequency, standing waves are produced. If the
end x = 0 is fixed and x = L is free, the equation is again given by:
y = 2A sin kx cos ωt$\omega t$
with the boundary condition that x = L is an antinode. The boundary condition that x = 0 is a node is automatically satisfied by the above equation. For x = L to be an antinode,

sinkL=±1$sinkL=\pm 1$

Or, kL=(n+12)π$kL=(n+{\displaystyle \frac{1}{2}})\pi$

Or, 2πLλ=(n+12)π${\displaystyle \frac{2\pi L}{\lambda }}=(n+{\displaystyle \frac{1}{2}})\pi$

Or, 2Lνv=n+12${\displaystyle \frac{2L\nu }{v}}=n+{\displaystyle \frac{1}{2}}$

Or, ν=(n+12)v2L=n+122LF/μ−−−−√$\nu =(n+{\displaystyle \frac{1}{2}}){\displaystyle \frac{v}{2L}}={\displaystyle \frac{n+{\displaystyle \frac{1}{2}}}{2L}}\sqrt{F/\mu }$.

These are the normal frequencies of vibration. The fundamental frequency is obtained when n=0, ie,

ν0=v/4L${\nu }_0=v/4L$.

Example

Write the equation of a standing wave in a tube

Example: Write the equation for the fundamental standing sound waves in a tube that is open at both ends. If the tube is 80 cm long and speed of the wave is 330 m/s. Represent the amplitude of the wave at an antinode by A.

Solution:
Since speed of sound wave=330m/s$330m/s$,ν=v2l=3302×0.8$\nu ={\displaystyle \frac{v}{2l}}={\displaystyle \frac{330}{2\times 0.8}}$
Thus ω=2πν=1297$\omega =2\pi \nu =1297$
For a wave, ωk=v${\displaystyle \frac{\omega }{k}}=v$⟹k=3.93$⟹k=3.93$
Equation of wave is y=Acos(kx)sin(ωt)$y=Acos(kx)sin(\omega t)$
Since tube is open at both ends, thus there is an antinode at x=0, and thus there should be the term: cos(kx)$cos(kx)$ instead of sin(kx)$sin(kx)$.

Example

State the parameters in the equation of a standing wave

The standard equation of a standing wave:
y=Aosinwtcoskx$y=A_{o}sinwtcoskx$

Where:
Ao$A_o$ : Amplitude
w$w$ : Angular frequency
k$k$ : Wave Number

Definition

Nodes and antinodes

A node is a point along a standing wave where the wave has minimum amplitude. For instance, in a vibrating guitar string, the ends of the string are nodes. By changing the position of the end node through frets, the guitarist changes the effective length of the vibrating string and thereby the note played. The opposite of a node is an anti-node, a point where the amplitude of the standing wave is a maximum. These occur midway between the nodes.

Definition

Understand resonance in standing waves in organ pipes with given driving frequency

The resonant frequencies of air columns depend upon the speed of sound in air as well as the length and geometry of the air column. Longitudinal pressure waves reflect from either closed or open ends to set up standing wave patterns. Important in the visualization of these standing waves is the location of the nodes and anti-nodes of pressure and displacement for the air in the columns.
Open- open tube : T=2Lvsound$T={\displaystyle \frac{2L}{v_{sound}}}$,    f=vsound2L$f={\displaystyle \frac{v_{sound}}{2L}}$

Close-close tube: T=2Lvsound$T={\displaystyle \frac{2L}{v_{sound}}}$,    f=vsound2L$f={\displaystyle \frac{v_{sound}}{2L}}$

Open-close tube : T=4Lvsound$T={\displaystyle \frac{4L}{v_{sound}}}$,    f=vsound4L$f={\displaystyle \frac{v_{sound}}{4L}}$

Example

Intensity at a certain distance from a isotropic point source delivering constant power

Example: Two coherent point sound sources S1$S_1$ and S2$S_2$ are placed as shown in the figure. Both are emitting sound of frequency 165 Hz$165Hz$.
S1$S_1$ is ahead of S2$S_2$ in phase by π−$\pi -$radian. (Speed of sound is 330 m/s$330m/s$)
If power of sources are same and equal to 7200$7200$ π$\pi$ watt. What will be resultant intensity (in watt/m2)$\left(inwatt/m^2\right)$ at point B.

Solution:
AB=82+62−−−−−−√=10$AB=\sqrt{8^2+6^2}=10$

Intensity due to S1$S_1$ at B: I1=7200π4π×102=18W/m2$I_1={\displaystyle \frac{7200\pi }{4\pi \times {10}^2}}=18W/m^2$

Intensity due to S2$S_2$ at B: I2=7200π4π×62=50W/m2$I_2={\displaystyle \frac{7200\pi }{4\pi \times 6^2}}=50W/m^2$

Wavelength λ=vf=330165=2m$\lambda ={\displaystyle \frac{v}{f}}={\displaystyle \frac{330}{165}}=2m$ as velocity and frequency are given.

The expression for intensity of two waves after interference is
I=I1+I2+2I1I2−−−−√cosϕ$I=I_1+I_2+2\sqrt{I_1{I}_2}\cos \varphi$  .....(1)
ϕ=π$\varphi =\pi$ as given.

Substituting ϕ$\varphi$, 

I=18+50+218×50−−−−−−√cosπ=18+50−218×50−−−−−−√=8 W/m2

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9  Page 10
Page 11  Page 12  Page 13  Page 14  Page 15