8. Beats : When two waves of same amplitude with slight difference in frequency (<10), traveling in the same direction superpose, beats are produced.
The equation for beats is
y =
ω −ω 2
)
2Acos( 1 2 t sin
ω −ω 2
2
1 t
Where amplitude at a given location
= 2Acos
ω −ω 2
2
1 t
The above expression shows that amplitude change with time.
Beat frequency = no of maxima / minima per second = v1 – v2
Waves & Doppler Effect
P HYSICS F UNDAMENTAL F OR IIT-J EE
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
XtraEdge for IIT-JEE 29 JANUARY 2011 9. Standing waves (stationary)
When two waves of same amplitude and frequency moving in opposite direction superimpose, standing waves are produced.
Nodes are the point where the displacement is always zero.
The amplitudes of different particles different and is maximum at antinodes.
The equation of standing waves is
The above expression shows that the amplitude is different for different values of x and varies sinusoidally.
For a node to occur at position x, y = 0 ⇒ kx = 0 For an antinode two occur at position x, y should
be max ⇒ kx = π/2 , ….
In terms of pressure ∆P = ∆P0 cos kx cos ωt.
10. For standing waves on strings (and both end open organ pipe)
Fundamental frequency v0 = λ0
v = 2l
v
First mode of vibration v1 = λ1
v = 2
2l
v
= 2v0 = 2nd harmonic
nth mode of vibration vn = n
2l
v = nv0
= nth harmonic where v = m
T for string.
Also more the tension in the same string, higher is the value of v0
11. For closed organ pipe : Fundamental frequency v0 =
λ0
v = 4l
v
First mode of vibration v1 = λ1
v = 3
4l
v = 3v0
= Third harmonic
nth mode of vibration vn = (2n + 1) 4l
v where n = 1, 2, …..
In case where end correction is taken replace l by (l + e)
12. (a) Intensity of sound at a distance r from a point source is I = 2
4 r P
π where P = power of source.
(b) For a line source I = rl P π where l is the length of source (c) I =
1 ρv(4π2 2v2)A2 =
2ρν
) amplitude essure
(Pr 2
13. Doppler's effect :
v = v0
±
±
s L
v v
v
v vL = velocity of listener
The above formula is valid when vs < v
Replace v by (v ± vm) if it is given that the medium also moving.
When listener and source are not moving along the line joining the two, then the component of velocity along the line joining the two are taken as velocity of listener or source.
14. If the source and listener are on the same vehicle and the sound is reflected from a stationary object towards which the vehicle is approaching then the frequency of sound as heard by the observer is v´ = v0
+ +
s L
v v
v v
15. For a path difference of λ, the phase difference is 2π for harmonic waves.
16. For a transverse wave the energy per unit length possessed by a string is given as
dl
dE = m(4π2f2)A2cos2 (kx – ωt)
17. Equation for a wave pulse is y = f(x + vt)
18. When a wave on reaching on interface is partly reflected and partly transmitted then for no power loss.
Pi = Pt + Pr where Pi = Power of incident wave Pt = Power of transmitted wave
Pr = Power of reflected wave.
Also in this case Ar =
+
−
1 2
1 2
v v
v
v Ai; At =
+ 2
1
2 2
v v
v Ai
Where Ai, Ar and At are amplitudes of incident reflected and transmitted waves v1 is the velocity in the medium of incidence and v2 is the velocity in the medium where transmitted wave is present.
Problem Solving Strategy : Mechanical Waves
Identify the relevant concepts : Wave problems fall into two broad categories. Kinematics problems are concerned with describing wave motion; they involve wave speed v, wave length λ(or wave number k), frequency f (or angular frequency ω), and amplitude A. They may also involve the position, velocity, and acceleration of individual particles in the medium.
Dynamics problems also use concepts from Newton's laws such as force and mass. In this chapter we'll encounter problems that involve the relation of wave speed to the mechanical properties of the wave medium. We'll get into these relations.
As always, make sure that you identify the target variable(s) for the problem. In some cases it will be the wavelength, frequency, or wave speed; in other
XtraEdge for IIT-JEE 30 JANUARY 2011 cases you'll be asked to find an expression for the
wave function.
Set up the problem using the following steps : Make a list of the quantities whose value are
given. To help you visualize the situation, you'll find it useful to sketch graphs of y versus x (fig.
a) and of y versus (fig. b). Label your graphs with the values of the known quantities.
y
Wave displacement versus coordinate x
at time t = 0
Wavelength λ A
A
(a)
y
Wave displacement versus time t at coordinate x = 0
Period T A
A
(b)
t t
Decide which equations you'll need to use. If any two of v, f, and λ are given, you'll need to use eq.
v = λf (periodic wave) to find the third quantity. If the problem involves the angular frequency ω and / or the wave number k, you'll need to use the definitions of those quantities and eq. (ω = vk).
You may also need the various forms of the wave function given in Eqs.
y(x, t) = A cos
−
ω t
v
x = A cos 2πf
−t v x ,
y(x, t) = A cos 2π
−
λ T
t
x
and y(x, t) = A cos (kx – ωt).
If the wave speed is not given, and you don't have enough information to determine it using v = λf, you may be able to find v using the relationship between v and the mechanical properties of the system.
Execute the solution as follows : Solve for the unknown quantities using the equations you've selected. In some problems all you need to do is find the value of one of the wave variables.
If you're asked to determine the wave function, you need to know A and any two of v, λ and f(or v, k and ω). Once you have this information, you can use it in eq. (ω = vk). You may also need the various forms of the wave function given in Eqs.
y(x, t) = A cos
−
ω t
v
x = A cos 2πf
−t v x ,
y(x, t) = A cos 2π
−
λ T
t
x and y(x, t) = A cos (kx – ωt) to get the specific wave function for the problem at hand. Once you have that, you can find the value of y at any point (value of x) and at any time by substituting into the wave function.
Evaluate your answer : Look at your results with a critical eye. Check to see whether the values of v, f, and λ (or v, ω, and k) agree with the relationships given in eq. . v = λf or w = vk. If you've calculated the wave function, check one or more special cases for which you can guess what the results ought to be.
Problem Solving Strategy : Standing waves
Identify the relevant concepts : As with traveling waves, it's useful to distinguish between the purely kinematic quantities, such as wave speed v, wavelength λ, and frequency f, and the dynamic quantities involving the properties of the medium, such as F and µ for transverse waves on a string.
Once you decide what the target variable is, try to determine whether the problem is only kinematic in nature or whether the properties of the medium are also involved.
Set up the problem using the following steps : In visualizing nodes and antinodes in standing
waves, it is always helpful to draw diagrams. For a string you can draw the shape at one instant and label the nodes N and antinodes A. The distance between two adjacent nodes or two adjacent antinodes is always λ/2, and the distance between a node and the adjacent antinode is always λ/4.
Decide which equation you'll need to use. The wave function for the standing wave is almost always useful ex. y(x, t) = (ASW sin kx) sin ωt.
You can compute the wave speed if you know either λ and f (or, equivalently, k = 2π/λ and ω = 2πf) or the properties of the medium (for a string. F and µ.)
Execute the solution as follows: Solve for the unknown quantities using the equations you've selected. Once you have the wave function, you can find the value of the displacement y at any point in the wave medium (value of x) and at any time. You can find the velocity of a particle in the wave medium by taking the partial derivative of y with respect to time. To find the acceleration of such a particle, take the second partial derivative of y with respect to time.
Evaluate your answer : Compare your numerical answers with your diagram. Check that the wave function is compatible with the boundary conditions (for example, the displacement should be zero at a fixed end).
Problem Solving Strategy : Sound Intensity
Identify the relevant concepts : The relationships between intensity and amplitude of a sound wave are rather straightforward. Quite a few other quantities are involved in these relationships, however, so it's particularly important to deciede which is your target variable.
Set up the problem using the following steps :
XtraEdge for IIT-JEE 31 JANUARY 2011 Sort the various physical quantities into
categories. The amplitude is described by A or pmax, and the frequency f can be determined from ω, k, or λ. These quantites are related through the wave speed v, which in turn is determined by the properties of the medium: B and ρ for a liquid; γ, T, and M for a gas.
Determine which quantities are given and which are the unknown target variables. Then start looking for relationships that take you where you want to go.
Execute the solution as follows: Use the equations you've selected to solve for the target variables. Be certain that all of the quantities are expressed in the correct units. In particular, if temperature is used to calculate the speed of sound in a gas, make sure that it is expressed in Kelvins (Celsius temperature plus 273.15).
Evaluate your answer: There are multiple relationships among the quantities that describe a wave. Try using an alternative one to check your results.
Problem Solving Strategy : Doppler Effect
Identify the relevant concepts : The Doppler effect is relevant whenever the source of waves, the wave detector (listener), or both are in motion.
Set up the problem using the following steps : Establish a coordinate system. Define the positive
direction to be the direction from the listener to the source, and make sure you know the signs of all relevant velocities. A velocity in the direction from the listener toward the source is positive; a velocity in the opposite direction is negative.
Also, the velocities must all be measured relative to the air in which the sound is traveling.
Use consistent notation to identify the various quantities: subscript S for source, L for listener.
Determine which unknown quantities are your target variables.
Execute the solutions : Use eq. fL = the source and the listener, the sound speed, and the velocities of the source and the listener. If the source is moving, you can find the wavelength measured by the listener using Eq.
λ = When a wave is reflected from a surface, either
stationary or moving, the analysis can be carried out in two steps. In the first, the surface plays the role of listener; the frequency with which the wave crests arrive at the surface is fL. Then think of the surface as a new source, emitting waves
with this same frequency fL. Finally, determine what frequency is heard by a listener detecting this new wave.
Evaluate your answer: Ask whether your final result makes sense. If the source and the listener are moving towards each other, fL > FS; if they are
(a) What are the amplitude and velocity of the component waves whose superposition can give rise to this vibration ?
(b) What is the distance between the nodes ?
(c) What is the velocity of a particle of the string at the position x = 1.5 cm when t = 9/8 s ?
Thus, the given stationary wave is formed by the superposition of the progressive waves
y1 =
Comparing each wave with the standard form of the progressive wave
y = a sin Distance between the nodes =
2
XtraEdge for IIT-JEE 32 JANUARY 2011
∴ At x = 1.5 cm and t = 8 9s v = – 200π sin 45π = 0
2. An engine blowing a whistle of frequency 133 Hz moves with a velocity of 60 m s–1 towards a hill from which an echo is heard. Calculate the frequency of the echo heard by the driver. (Velocity of sound in air
= 340 ms–1.)
Sol. The 'image' of the source approaches the driver at the same speed. Here, the image or echo is the source.
∴ vs = + 60 ms–1, v0 = – 60 ms–1
3. A source of sound of frequency 1000 Hz moves to the right with a speed of 32 ms–1 relative to the ground. To its right is reflecting surface moving to the left with a speed of 64 ms–1 relative to the ground.
Take the speed of sound in air to be 332 m s–1 and find
(a) the wavelength of the sound emitted in air by the source
(b) the number of waves per second arriving at the reflecting surface
(c) the speed of the reflected waves, and (d) the wavelength of the reflected waves
Sol. (a) Due to the motion of the source, the wavelength (and hence, the frequency) is actually changed from λ to λ´ such that if n = actual frequency
(b) The number of waves arriving at the reflecting surface is the same as the number of waves received by an observer moving towards the source. This is given by the apparent frequency.
n´ = (c) Same as that of the incident wave because the
speed of a wave depends only on the characteristics of the medium.
∴ speed of the reflected wave = 332 ms–1
(d) To calculate the wavelength of the reflected wave, we may consider the source to be stationary and emitting waves of wavelength 0.3 m. If the reflector were stationary, waves in a tube of length c would reach the reflector and the same number of reflected waves would be contained in a tube of the same length, so the wavelength of the reflected wave would also be the same as that of the incident wave. But when the reflector moves towards the source with speed vref´ it would reflect additional waves contained in vref and the total number of waves reflected would be
contained in a tube of length c – vref. If λ´is the changed wavelength of the wave due to the motion of the source
λ ´´ =
4. Find the ratio of the fundamental frequencies of two identical strings after one of them is stretched by 2%
and the other by 4%. tension is proportional to the increase in length, T = k × fl0 where k is a constant.
5. An open organ pipe has a fundamental frequency of 300 Hz. The first overtone of a closed organ pipe has the same frequency as the first overtone of the open pipe. How long is each pipe ? The velocity of sound in air = 350 ms–1. ... are the overtones.
⇒ l = 2cn = 300 2
350
× = 0.58 m The frequency of the first overtone = 2n = 2 × 300 = 600 Hz
∴ the frequency of the first overtone of the closed pipe = 600 = 3n
XtraEdge for IIT-JEE 33 JANUARY 2011 Definition and Classification :
Carbohydrates are polyhydroxy aldehydes, polyhydroxy ketones, or compounds that can be hydrolyzed to them. A carbohydrate that cannot be hydrolyzed to simpler compounds is called a monosaccharide. A carbohydrate that can be hydrolyzed to two monosaccharide molecules is called a disaccharide. A carbohydrate that can be hydrolyzed to many monosaccharide molecules is called a polysaccharide.
A monosaccharide may be further classified. If it contains an aldehyde group, it is known as an aldose;
if it contains a keto group, it is known as a ketose.
Depending upon the number of carbon atoms. It contains, a monosaccharide is known as a triose, tetrose, pentose, hexose, and so on. An aldohexose, for example, is a six-carbon monosaccharide containing an aldehyde group; a ketopentose is a five-carbon monosaccharide containing a keto group.
Most naturally occurring monosaccharides are pentoses or hexoses.
Carbohydrates that reduce Fehling’s (or Benedict’s) or Tollens’ reagent are known as reducing sugars.
All monosaccharides, whether aldose or ketose, are known as reducing sugars. Most disaccharides are reducing sugars; sucrose (common table sugar) is a notable exception, for it is a non-reducing sugar.
(+)-Glucose : an aldohexose :
Because it is the unit of which starch, cellulose, and glycogen are made up, and because of its special role in biological processes, (+)-glucose is by far the most abundant monosaccharide- there are probably more (+)-glucose units in nature than any other organic group–and by far the most important monosaccharide.
Cyclic structure of D-(+)-glucose. Formation of glucosides :
glucose is a pentahydroxy aldehyde. D-(+)-glucose had been definitely proved to have structure.
CHO OH H
OH OH H
HO H H
CH2OH D-(+)-Glucose
By 1895 it had become clear that the picture of D-(+)-glucose as a pentahydroxy aldehyde had to be modified.
Among the facts that had still to be accounted for were the following:
(a) D-(+)-Glucose fails to undergo certain reactions typical of aldehydes. Although it is readily oxidized, it gives a negative Schiff test and does not form a bisulfite addition product.
(b) D-(+)-Glucose exists in two isomeric forms which undergo mutarotation. When crystals of ordinary D-(+)-glucose of m.p. 146ºC are dissolved in water, the specific rotation gradually drops from an initial + 112º to + 52.7º. On the other hand, when crystals of D-(+)-glucose of m.p. 150ºC (obtained by crystallization at temperatures above 98ºC) are dissolved in water, the specific rotation gradually rises from an initial + 19º to + 52.7º. The form with the higher positive rotation is called α-D-(+)-glucose and that with lower rotation β-D-(+)-glucose. The change in rotation of each of these to the equilibrium value is called mutarotation.
(c) (+)-Glucose forms two isomeric methyl D-glucosides. Aldehydes react with alcohols in the presence of anhydrous HCl to form acetals. If the alcohol is, say methanol, the acetal contains two methyl groups :
–C=O H
–C–OCH3
H
OH
–C–OCH3
H
OCH3 CH3OH,H+ CH3OH,H+
Aldehyde Hemiacetal Acetal When D-(+)-glucose is treated with methanol and HCl, the product, methyl D-glucoside, contains only one –CH3 group; yet it has properties resembling those of a full acetal. It does not spontaneously revert to aldehyde and alcohol on contact with water, but requires hydrolysis by aqueous acids.
Furthermore, not just one but two of these monomethyl derivatives of D-(+)-glucose are known, one with m.p. 165ºC and specific rotation + 158º, and the other with m.p. 107 ºC and specific rotation –33º.
The isomer of higher positive rotation is called methyl α-D-glucoside, and the other is called methyl β-D-glucoside. These glucosides do not undergo mutarotation, and do not reduce Tollens’ or Fehling’s reagent.