 Physics BEATS AND DOPPLER EFFECT

### Topic Covered

color{blue}{star} Standing Waves and Normal Modes
color{blue}{star} BEATS
color{blue}{star} DOPPLER EFFECT

### Standing Waves and Normal Modes

color{blue} ✍️ We considered above reflection at one boundary. But there are familiar situations (a string fixed at either end or an air column in a pipe with either end closed) in which reflection takes place at two or more boundaries.

color{blue} ✍️ In a string, for example, a wave going to the right will get reflected at one end, which in turn will travel and get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called "standing waves" or stationary waves.

color{blue} ✍️To see this mathematically, consider a wave travelling along the positive direction of x-axis and a reflected wave of the same amplitude and wavelength in the negative direction of x-axis. From Eqs. (15.2) and (15.4), with φ = 0, we get:

color{ blue } {y_1(x, t) = a sin (kx – ωt)}
color{blue} {y_2(x, t) = a sin (kx + ωt)}

color{blue} ✍️The resultant wave on the string is, according to the principle of superposition:

color {brown} { y (x, t) = y_1(x, t) + y_2(x, t)}

color {purple} { = a [sin (kx – ωt) + sin (kx + ωt)]}

color{blue} ✍️Using the familiar trignometric identity

color{purple} {Sin (A+B) + Sin (A–B) = 2 sin A cosB} we get,

color{blue} { y (x, t) = 2a sin kx cos ωt}

........................ (15.37)

color{brown} {"Note"} the important difference in the wave pattern described by Eq. (15.37) from that described by Eq. (15.2) or Eq. (15.4).

color{blue} ✍️The terms kx and ωt appear separately, not in the combination kx - ωt. The amplitude of this wave is 2a sin kx.

color{blue} ✍️Thus in this wave pattern, the amplitude varies from point to point, but each element of the string oscillates with the same angular frequency ω or time period. There is no phase difference between oscillations of different elements of the wave.

color{blue} ✍️The string as a whole vibrates in phase with differing amplitudes at different points. The wave pattern is neither moving to the right nor to the left. Hence they are called "standing or stationary waves."

color{blue} ✍️The amplitude is fixed at a given location but, as remarked earlier, it is different at different locations. The points at which the amplitude is zero (i.e., where there is no motion at all) are nodes; the points at which the amplitude is the largest are called "antinodes."

color{blue} ✍️Fig. 15.12 shows a stationary wave pattern resulting from superposition of two travelling waves in opposite directions. color{blue} ✍️The most significant feature of stationary waves is that the boundary conditions constrain the possible wavelengths or frequencies of vibration of the system.

color{blue} ✍️The system cannot oscillate with any arbitrary frequency (contrast this with a harmonic travelling wave), but is characterized by a set of natural frequencies or normal modes of oscillation. Let us determine these normal modes for a stretched string fixed at both ends.

color{blue} ✍️First, from Eq. (15.37), the positions of nodes (where the amplitude is zero) are given by sin kx = 0 . which implies

kx = nπ; n = 0, 1, 2, 3, ...

Since k = 2π//λ , we get

color{blue} {x = (nλ)/2 ; n = 0, 1, 2, 3, ...}

....................... (15.38)

color{blue} ✍️Clearly, the distance between any two successive nodes is lamda/2 . In the same way, the positions of antinodes (where the amplitude is the largest) are given by the largest value of sin kx :

|sin k x|= 1

color{blue} ✍️ which implies color {purple} {kx = (n + ½) π ; n = 0, 1, 2, 3, ...}

color{blue} ✍️With k = 2π//λ, we get

color { blue} { x = (n + ½) lamda/2 ; n = 0, 1, 2, 3, ...}

................................ (15.39)

color{blue} ✍️Again the distance between any two consecutive antinodes is lamda/2. Eq. (15.38) can be applied to the case of a stretched string of length L fixed at both ends.

color{blue} ✍️Taking one end to be at x = 0, the boundary conditions are that x = 0 and x = L are positions of nodes. The x = 0 condition is already satisfied. The x = L node condition requires that the length L is related to λ by

color{ blue} { L = (n lamda)/2 ; n = 1, 2, 3, ...}

................... (15.40)

color{blue} ✍️Thus, the possible wavelengths of stationary waves are constrained by the relation

color{blue} {lamda = (2L)/n ; n = 1, 2, 3, …}

..................... (15.41)

color{blue} {nu = (n nu )/(2L) ," for "n = 1, 2, 3},

.......................(15.42)

color{blue} ✍️We have thus obtained the natural frequencies - the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic.

color{blue} ✍️For the stretched string fixed at either end it is given by nu = nu/(2L), corresponding to n = 1 of Eq. (15.42).

color{blue} ✍️Here v is the speed of wave determined by the properties of the medium. The n = 2 frequency is called the second harmonic; n = 3 is the third harmonic and so on. We can label the various harmonics by the symbol νn \ \ ( n = 1, 2, ...).

color{blue} ✍️Fig. 15.13 shows the first six harmonics of a stretched string fixed at either end. A string need not vibrate in one of these modes only.

color{blue} ✍️Generally, the vibration of a string will be a superposition of different modes; some modes may be more strongly excited and some less. Musical instruments like sitar or violin are based on this principle. Where the string is plucked or bowed, determines which modes are more prominent than others. color{blue} ✍️Let us next consider normal modes of oscillation of an air column with one end closed and the other open. A glass tube partially filled with water illustrates this system.

color{blue} ✍️The end in contact with water is a node, while the open end is an antinode. At the node the pressure changes are the largest, while the displacement is minimum (zero).

color{blue} ✍️At the open end - the antinode, it is just the other way - least pressure change and maximum amplitude of displacement. Taking the end in contact with water to be x = 0, the node condition (Eq. 15.38) is already satisfied. If the other end x = L is an antinode, Eq. (15.39) gives

color{blue} { L = ( n + 1/2 ) lamda/2 }" for "n = 0, 1, 2, 3, …

color{blue} ✍️The possible wavelengths are then restricted by the relation :

color{blue} {lamda = ( 2L)/ (( n + 1/2 )) ," for "n = 0, 1, 2, 3,... }

.......................(15.43)

color{blue} ✍️The normal modes – the natural frequencies – of the system are

color{blue} {nu = ( n + 1/2 ) nu /(2L) ; n = 0, 1, 2, 3, }

................ (15.44)

color{blue} ✍️The fundamental frequency corresponds to n = 0, and is given by nu/(4L) . The higher frequencies are odd harmonics, i.e., odd multiples of the fundamental frequency : 3 nu/(4L) , 5 nu/(4L) , etc.

color{blue} ✍️Fig. 15.14 shows the first six odd harmonics of air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode.

color{blue} ✍️It is then easily seen that an open air column at both ends generates all harmonics (See Fig. 15.15). The systems above, strings and air columns, can also undergo forced oscillations. If the external frequency is close to one of the natural frequencies, the system shows resonance.  color{blue} ✍️Normal modes of a circular membrane rigidly clamped to the circumference as in a tabla are determined by the boundary condition that no point on the circumference of the membrane vibrates.

color{blue} ✍️Estimation of the frequencies of normal modes of this system is more complex. This problem involves wave propagation in two dimensions.
Q 3169778615 A pipe, 30.0 cm long, is open at both ends. Which harmonic mode of the pipe resonates a 1.1 kHz source? Will
resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as 330  m s^-1.
Class 11 Chapter 15 Example 5 Solution:

The first harmonic frequency is given by

nu_1 = nu/(lamda_1 = nu/(2L) (open pipe )

where L is the length of the pipe. The frequency of its nth harmonic is:

V_n = ( n nu )/ (2L), for n = 1, 2, 3, ... (open pipe)

First few modes of an open pipe are shown in Fig15.14

For L = 30.0 cm, v = 330 m s^-1,

v_n = ( n xx 330 (ms^-1) ) / (0.6 (m)) = 550 ns^-1

Clearly, a source of frequency 1.1 kHz will resonate at v_2, i.e. the second harmonic.

Now if one end of the pipe is closed (Fig. 15.15), it follows from Eq. (14.50) that the fundamental frequency is

v_1 = nu /lamda_1 = nu/(4L) (pipe closed at one end)

and only the odd numbered harmonics are present :

v _3 = (3 nu ) /( 4L) , v_5 = (5nu ) / ( 4L) , and so on

For L = 30 cm and v = 330 m s^-1, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency
corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

### BEATS

color{blue} ✍️‘Beats’ is an interesting phenomenon arising from interference of waves. When two harmonic sound waves of close (but not equal) frequencies are heard at the same time, we hear a sound of similar frequency (the average of two close frequencies), but we hear something else also.

color{blue} ✍️We hear audibly distinct waxing and waning of the intensity of the sound, with a frequency equal to the difference in the two close frequencies. Artists use this phenomenon often while tuning their instruments with each other. They go on tuning until their sensitive ears do not detect any beats.

color{blue} ✍️To see this mathematically, let us consider two harmonic sound waves of nearly equal angular frequency ω1 and ω2 and fix the location to be x = 0 for convenience. Eq. (15.2) with a suitable choice of phase (φ = π//2 for each) and, assuming equal amplitudes, gives

color{blue} {s_1 = a cos ω_1t} and color{blue} {s_2 = a cos ω_2t}

.......................... (15.45)

color{blue} ✍️Here we have replaced the symbol y by s, since we are referring to longitudinal not transverse displacement. Let ω_1 be the (slightly) greater of the two frequencies. The resultant displacement is, by the principle of superposition,

 color{green} {s = s_1 + s_2 = a (cos ω_1 t + cos ω_2 t)}

color{blue} ✍️Using the familiar trignometric identity for cos A + cosB, we get

color{green} {=2 a cos ((omega_1 - omega_2)t)/2 cos ((omega_1 + omega_2)t)/2}

.......15.46

color{blue} ✍️which may be written as :

color{blue} {s = [2 a cos ω_b t ] cos ω_at }

.............(15.47)

color{blue} ✍️If |ω_1 – ω_2| < < ω_1, ω_2, ω_a > > ω_b,

where

color{blue} {omega_b = ((omega_1 - omega_2))/2 } and color{blue} {omega_a = (omega_1 + omega_2)/2}

color{blue} ✍️Now if we assume | omega_1 - omega_2 | < < omega_1 , which means ω_a > > ω_b, we can interpret Eq. (15.47) as follows.

color{blue} ✍️The resultant wave is oscillating with the average angular frequency ω_a; however its amplitude is not constant in time, unlike a pure harmonic wave. The amplitude is the largest when the term cos ω_b t takes its limit +1 or –1.

color{blue} ✍️ In other words, the intensity of the resultant wave waxes and wanes with a frequency which is 2ω_b = ω_1 - omega_2 Since ω = 2πν, the beat frequency ν_"beat", is given by

color{blue} {ν_"beat" = ν_1 – ν_2}

........................ (15.48)

color{blue} ✍️Fig. 15.16 illustrates the phenomenon of beats for two harmonic waves of frequencies 11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency of 2 Hz. Q 3159178914 Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz ?
Class 11 Chapter 15 Example 6 Solution:

Increase in the tension of a string increases its frequency. If the original frequency of B (ν_B) were greater than that of A (ν_A ), further increase in ν_B should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that ν_B < ν_A. Since ν_A – ν_B = 5 Hz, and ν_A = 427 Hz, we get ν_B = 422 Hz.

### DOPPLER EFFECT

color{blue} ✍️It is an everyday experience that the pitch (or frequency) of the whistle of a fast moving train decreases as it recedes away. When we approach a stationary source of sound with high speed, the pitch of the sound heard appears to be higher than that of the source.

color{blue} ✍️As the observer recedes away from the source, the observed pitch (or frequency) becomes lower than that of the source. This motion-related frequency change is called color{blue} "Doppler effect".

color{blue} ✍️The Austrian physicist Johann Christian Doppler first proposed the effect in 1842. Buys Ballot in Holland tested it experimentally in 1845. Doppler effect is a wave phenomenon, it holds not only for sound waves but also for electromagnetic waves. However, here we shall consider only sound waves.

color{blue} ✍️We shall analyse changes in frequency under three different situations:

color{blue} {(1)} observer is stationary but the source is moving,
color{blue} {(2)} observer is moving but the source is stationary, and
color{blue} {(3)} both the observer and the source are moving.

color{blue} ✍️The situations (1) and (2) differ from each other because of the absence or presence of relative motion between the observer and the medium. Most waves require a medium for their propagation; however, electromagnetic waves do not require any medium for propagation.

color{blue} ✍️ If there is no medium present, the Doppler shifts are same irrespective of whether the source moves or the observer moves, since there is no way of distinction between the two situations.

### Source Moving ; Observer Stationary

color{blue} ✍️Let us choose the convention to take the direction from the observer to the source as the positive direction of velocity. Consider a source S moving with velocity vs and an observer who is stationary in a frame in which the medium is also at rest.

color{blue} ✍️Let the speed of a wave of angular frequency ω and period T_o, both measured by an observer at rest with respect to the medium, be v. We assume that the observer has a detector that counts every time a wave crest reaches it.

color{blue} ✍️As shown in Fig. 15.17, at time t = 0 the source is at point S_1, located at a distance L from the observer, and emits a crest. This reaches the observer at time t_1 = L//v. At time t = T_o the source has moved a distance v_sT_o and is at point S_2, located at a distance (L + v_sT_o) from the observer. At S_2, the source emits a second crest.

color{blue} ✍️This reaches the observer at t_2 = T_0 + (L + v_sT_0)/nu color{blue} ✍️At time n T_o, the source emits its (n+1)^(th) crest and this reaches the observer at time

color{brown} { t_(n +1) = n T_0 + (( L + n v_s T_0))/nu}

color{blue} ✍️Hence, in a time interval

color{brown} { [nT_0 + ((L + nv_sT_0 ))/nu - L/nu]}

color{blue} ✍️the observer’s detector counts n crests and the observer records the period of the wave as T given by

T = [nT_0 + ((L + nv_sT_0 ))/nu - L/nu]//n

color{purple} {T= T_0 + (nu_sT_o)/nu}

color {blue} {= T_0 ( 1 + nu_s/nu )}

..........................(15.49)

color{blue} ✍️Equation (15.49) may be rewritten in terms of the frequency v_o that would be measured if the source and observer were stationary, and the frequency v observed when the source is moving, as

color{blue} {nu = nu_0 ( 1 + v_s/nu )^-1}

................(15.50)

color{blue} ✍️If v_s is small compared with the wave speed v, taking binomial expansion to terms in first order in v_s//v and neglecting higher power, Eq. (15.50) may be approximated, giving

color{blue} { nu = nu_0( 1 - nu_s/nu )}

........(15.51)

color{blue} ✍️For a source approaching the observer, we replace v_s by – v_s to get

color{blue} { nu = nu_0( 1 + nu_s/nu )}

..................(15.52)

color{blue} ✍️The observer thus measures a lower frequency when the source recedes from him than he does when it is at rest. He measures a higher frequency when the source approaches him.

### Observer Moving; Source Stationary

color{blue} ✍️Now to derive the Doppler shift when the observer is moving with velocity v_o towards the source and the source is at rest, we have to proceed in a different manner.

color{blue} ✍️We work in the reference frame of the moving observer. In this reference frame the source and medium are approaching at speed v_o and the speed with which the wave approaches is v_o + v.

color{blue} ✍️Following a similar procedure as in the previous case, we find that the time interval between the arrival of the first and the (n+1)^( th) crests is

color{brown} { t_(n+1) - t_1 = n T_0 - (n nu_0T_0)/ (nu_0 + nu)}

color{blue} ✍️The observer thus, measures the period of the wave to be

= T_0 ( 1 - nu_0/(nu_0 + nu ) )

= T_0 ( 1 + nu_0/nu )^-1

giving

color{blue} {nu = nu_0 ( 1 + nu_0/nu)}

.................(15.53)

color{blue} ✍️If nu_0/nu is small , the Doppler shift is almost same whether it is the observer or the source moving since Eq. (15.53) and the approximate relation Eq. (15.51 ) are the same.

### Both Source and Observer Moving

color{blue} ✍️We will now derive a general expression for Doppler shift when both the source and the observer are moving. As before, let us take the direction from the observer to the source as the positive direction.

color{blue} ✍️Let the source and the observer be moving with velocities v_s and v_o respectively as shown in Fig.15.18. Suppose at time t = 0, the observer is at O_1 and the source is at S_1, O_1 being to the left of S_1. color{blue} ✍️The source emits a wave of velocity v, of frequency v and period T_0 all measured by an observer at rest with respect to the medium. Let L be the distance between O_1 and S_1 at t = 0, when the source emits the first crest. Now, since the observer is moving, the velocity of the wave relative to the observer is v +v_0.

color{blue} ✍️Therefore, the first crest reaches the observer at time t_1 = L/ (v+v_0). At time t = T_0, both the observer and the source have moved to their new positions O_2 and S_2 respectively. The new distance between the observer and the source, O_2 S_2, would be L+(v_s– v_0) T_0]. At S_2, the source emits a second crest.

color{brown}bbul{"Application of Doppler effect"}
color{blue} ✍️ The change in frequency caused by a moving object due to Doppler effect is used to measure their velocities in diverse areas such as military, medical science, astrophysics, etc. It is also used by police to check over-speeding of vehicles.

color{blue} ✍️ A sound wave or electromagnetic wave of known frequency is sent towards a moving object. Some part of the wave is reflected from the object and its frequency is detected by the monitoring station. This change in frequency is called Doppler shift.

color{blue} ✍️ It is used at airports to guide aircraft, and in the military to detect enemy aircraft. Astrophysicists use it to measure the velocities of stars.

color{blue} ✍️ Doctors use it to study heart beats and blood flow in different parts of the body. Here they use ulltrasonic waves, and in common practice, it is called sonography. Ultrasonic waves enter the body of the person, some of them are reflected back, and give information about motion of blood and pulsation of heart valves, as well as pulsation of the heart of the foetus. In the case of heart, the picture generated is called echocardiogram.

color{blue} ✍️This reaches the observer at time.

color{brown} { t_2 = T_o + ([L + (v_s – v_o)T_o ]) /(v + v_o)}

color{blue} ✍️At time nT_o the source emits its (n+1)^(th) crest and this reaches the observer at time

color{blue} { t_(n+1) = nT_o + ([L + n (v_s – v_o)T_o]) /(v + v_o )}

,

color{blue} ✍️Hence, in a time interval t_(n+1) –t_1, i.e.,

color{blue} { nT_o + ([L + n (v_s – v_o)T_o]) /(v + v_o ) – L /(v + v_o )},

color{blue} ✍️the observer counts n crests and the observer records the period of the wave as equal to T given by

color{blue} {T = T_0 ( 1 + (v_s - v_0 )/(v + v_0 ) ) = T_0 ((v +v_s)/(v- v_0 ))}

.......................(15.54)

color{blue} ✍️The frequency v observed by the observer is given by

color{blue} {v = v_0 ((v + v_0)/(v + v_s ))}

............. (15.55)

color{blue} ✍️Consider a passenger sitting in a train moving on a straight track. Suppose she hears a whistle sounded by the driver of the train. What frequency will she measure or hear? Here both the observer and the source are moving with the same velocity, so there will be no shift in frequency and the passenger will note the natural frequency.

color{blue} ✍️But an observer outside who is stationary with respect to the track will note a higher frequency if the train is approaching him and a lower frequency when it recedes from him.

color{brown} {"Note"} that we have defined the direction from the observer to the source as the positive direction. Therefore, if the observer is moving towards the source, v_0 has a positive (numerical) value whereas if O is moving away from S, v_0  has a negative value.

color{blue} ✍️On the other hand, if S is moving away from O, v_s has a positive value whereas if it is moving towards O, v_s has a negative value. The sound emitted by the source travels in all directions. It is that part of sound coming towards the observer which the observer receives and detects. Therefore, the relative velocity of sound with respect to the observer is v + v_0 in all cases.
Q 3159380214 A rocket is moving at a speed of 200 m s^-1 towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket.
Class 11 Chapter 15 Example 7 Solution:

(1) The observer is at rest and the source is moving with a speed of 200 m s^-1. Since this is comparable with the velocity of sound, 330 m s^-1, we must use Eq. (15.50) and not the approximate Eq. (15.51). Since the source is approaching a stationary target, v_o = 0, and v_s must be replaced by –v_s. Thus, we have

v = v_0 ( 1 - v_s/v ) ^-1

v = 1000 Hz × [1 – 200 m s–1/330 m s^-1]^-1

= 2540 Hz

(2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, v_s = 0 and vo has a positive value. The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not v_o. Therefore, the frequency as registered by the rocket is

v' = v ((v + v_0 )/v)

= 2540 Hz xx ( (200 ms^-1 + 330 ms^-1)/(330 ms^-1))

~~ 4080 Hz 