Sound Waves Concept Page - 8

Formula

Intensity

Example: The intensity of sound wave whose frequency is 250Hz$250Hz$ is π2×10−9xW/m2${\displaystyle {\displaystyle \frac{{\pi }^2\times {10}^{-9}}{x}}W/m^2}$. The displacement amplitude of particles of the medium at this position is 1×10−8m$1\times {10}^{-8}m$. The density of the medium is 1kg/m3$1kg/m^3$, bulk modulus of elasticity of the medium is 400N/m2$400N/m^2$. Find x$x$

Solution:
Given:    ν=250Hzρ=1kg/m3so=1×10−8m$\nu =250Hz\rho =1kg/m^3{s}_o=1\times {10}^{-8}m$Bulk modulus of elasticity     B=400N/m2$B=400N/m^2$Velocity of the sound in the medium        v=Bρ−−√=4001−−−−√m/s$v=\sqrt{{\displaystyle \frac{B}{\rho }}}=\sqrt{{\displaystyle \frac{400}{1}}}m/s$⟹v=20m/s$⟹v=20m/s$  Intensity of sound wave      I=2π2ν2s2oρv$I=2{\pi }^2{\nu }^2{s}_o^2\rho v$
I=2π2×(250)2×(10−8)2×1×20$I=2{\pi }^2\times (250{)}^2\times ({10}^{-8}{)}^2\times 1\times 20$
⟹I=0.25π2×10−9W/m2$⟹I=0.25{\pi }^2\times {10}^{-9}W/m^2$
Thus x=10.25=4$x={\displaystyle \frac{1}{0.25}}=4$

Intensity(I$I$) of a wave at a point of the medium is measured as the amount of sound energy passing per second normally through unit area at that point. Its unit is Wm−2$W{m}^{-2}$. Minimum intensity of sound wave audible to human ears is 10−12 Wm−2${10}^{-12}W{m}^{-2}$. Also, I∝A2$I\propto A^2$ where, A$A$ is the amplitude.

Result

Factors affecting intensity

Factors affecting intensity are the amplitude of vibration, density of the medium and the distance of source from human ear.

Example

Variation of Intensity with amplitude of wave

Example: If the amplitude of sound is doubled and the frequency is reduced to one-fourth, then find the intensity of sound at the same point.

Solution:
Given : f1f2=11/4=41${\displaystyle \frac{f_1}{f_2}}={\displaystyle \frac{1}{{}^1{/}_4}}={\displaystyle \frac{4}{1}}$
A1A2=12${\displaystyle \frac{A_1}{A_2}}={\displaystyle \frac{1}{2}}$
I=2π2f2A2ρv$I=2{\pi }^2{f}^2{A}^2\rho v$
I1I2=2π2f21A21ρv2π2f22A22ρv${\displaystyle \frac{I_1}{I_2}}={\displaystyle \frac{2{\pi }^2{f}_1^2{A}_1^2\rho v}{2{\pi }^2{f}_2^2{A}_2^2\rho v}}$
=f21A21f22A22$={\displaystyle \frac{f_1^2{A}_1^2}{f_2^2{A}_2^2}}$
=(f1f2)2(A1A2)2$={\left({\displaystyle \frac{f_1}{f_2}}\right)}^2{\left({\displaystyle \frac{A_1}{A_2}}\right)}^2$
=(4)2(12)2$=(4{)}^2{\left({\displaystyle \frac{1}{2}}\right)}^2$
=16×14$=16\times {\displaystyle \frac{1}{4}}$
=4$=4$
So, I1I2=4$So,{\displaystyle \frac{I_1}{I_2}}=4$
⇒I2=I14$\Rightarrow I_2={\displaystyle \frac{I_1}{4}}$
So, the intensity is decreased to 14th${{\displaystyle \frac{1}{4}}}^{th}$

Example

Numerical - Decibel level of Sound

Calculate the decibel level of : Here Io=1012wm2$I_o={10}^{12}{\displaystyle \frac{w}{m^2}}$
T.V. (Average Volume from 10 feet) - where
I=320×107wm2$I=320\times {10}^7{\displaystyle \frac{w}{m^2}}$
So D=10logIIo$D=10log{\displaystyle \frac{I}{I_o}}$
putting all this values we get 
D = 2.5

Example

Find nodes and antinodes for a standing waves in a string for different overtones

A sonometer wire is vibrating in the third overtone. There are how many no. of nodes and antinodes?

Solution:
The third overtone frequency
ν3=7v4L=7νo${\nu }_3=\frac{7v}{4L}=7{\nu }_o$
The third overtone has (n+1) Antinodes and (n+2) nodes,
i.e 4 Antinodes and 5 nodes.

Definition

Possible wavelengths in a pipe closed at one end

First harmonic : λ=4L$\lambda =4L$
There is no such thing as a 2nd harmonic for closed end pipes. In fact, all of the harmonics in closed end pipes are going to be odd numbers.
Third harmonic : λ=4L3$\lambda ={\displaystyle \frac{4L}{3}}$
Fifth harmonic : There is one full wavelength in there (4/4) plus an extra 14${\displaystyle \frac{1}{4}}$ of a wavelength for a total of 5/4. λ=4L5$\lambda ={\displaystyle \frac{4L}{5}}$

Example

Problem on pipe closed at one end at resonance

Example: A pipe closed at one end and open at the other end resonates with a sound of frequency 135 Hz$135Hz$ and also with 165 Hz$165Hz$, but not at any other frequency intermediate between these two. Then, find the frequency of the fundamental note of the pipe.

Solution:
Frequency of pipe closed at one end is given as: (2n+1)v4l${\displaystyle \frac{(2n+1)v}{4l}}$
Here, 165135=2n2+12n1+1${\displaystyle \frac{165}{135}}={\displaystyle \frac{2n_2+1}{2n_1+1}}$ --------- {1}
and n2=n1+1$n_2=n_1+1$
Hence, by substituting above equation in {1} and solving we get, n1=4$n_1=4$.Thus, n2=5$n_2=5$Thus, 165=(2n2+1)v2l$165={\displaystyle \frac{(2n_2+1)v}{2l}}$Thus, fundamental frequency = v2l=15 Hz${\displaystyle \frac{v}{2l}}=15Hz$

Example

Find the antinodes of a given sinusoidal standing wave in a pipe closed at one end

Example: In closed pipes, find the positions of antinodes  obtained.

Solution:
In closed pipe frequency is given by, f=(2n−1)v4L$f={\displaystyle \frac{(2n-1)v}{4L}}$
Where, n=$n=$ integer, v=$v=$ speed of sound, L=$L=$ length of the pipePosition of antinodes is given by (2n−1)λ4${\displaystyle \frac{(2n-1)\lambda }{4}}$Putting value of n$n$ as 0,1,2$0,1,2$ we get, λ4,3λ4${\displaystyle \frac{\lambda }{4}},{\displaystyle \frac{3\lambda }{4}}$ and 4λ4${\displaystyle \frac{4\lambda }{4}}$

Definition

Fundamental mode or first harmonic for standing wave in a pipe closed at one end

In an open organ pipe, the fundamental frequency is 30Hz.For an open pipe closed at one end fundamental frequency νo=v2L=30Hz${\nu }_o={\displaystyle \frac{v}{2L}}=30Hz$     -----(1)
In an closed pipe of same length fundamental frequency
νc=v4L=12(v2L)=12×30${\nu }_c={\displaystyle \frac{v}{4L}}={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{v}{2L}}\right)={\displaystyle \frac{1}{2}}\times 30$     ( From (1)  )
= 15 Hz

Example

Higher harmonics and overtones for standing wave in a pipe closed at one end

Example: If the fundamental frequency of a pipe closed at one is 512 Hz. Find the frequency of a pipe of the same dimension but open at both ends.

Solution:
At fundamental frequency of closed organ pipe, the open end will have an antinode and closed end will have a node λ4=L${\displaystyle \frac{\lambda }{4}}=L$, where L$L$ is length of air column. 
f=cλ$f={\displaystyle \frac{c}{\lambda }}$ =c4L=512 Hz$={\displaystyle \frac{c}{4L}}=512Hz$
At fundamental frequency of open organ pipe, both the open end will have an  antinode λ2=L${\displaystyle \frac{\lambda }{2}}=L$, where L$L$ is length of air column.f=cλ$f={\displaystyle \frac{c}{\lambda }}$ =c2L$={\displaystyle \frac{c}{2L}}$  =2×c4L=1024 Hz

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