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Velocity of sound in air at standard temperature and pressure is 332 m s^-1.

Wave velocity υ = nλ; where n = frequency, λ = wavelength.

(a) F 

(b) wavelength 

(c) 50 m

(i) C for compressions and R for rarefactions. Wavelength of longitudinal wave The distance between corresponding points of two consecutive compressions or two consecutive rarefactions is the wavelength of the longitudinal wave.

(ii) Wavelength = 1 m

(iii) Velocity V = f λ = 92 × 1 = 92 m/s

The time to complete one the vibration called Time period.

It is a non-mechanical wave because it does not require a medium. These waves travel from ground to satellites and then relayed back to our antenna. There is no medium in the space near the satellite.

The motion of air is not a wave motion because in a wave motion the particles of the medium do not move from one place to another but vibrate at its place.

The displacement of a particle on the string is given as 

y = A.sin[⍵(t-x/v) + ⲫ] 

where ⲫ is the phase constant.  

Since a value of the phase constant will be unique for one oscillation and it will be repeated at each 2π angle. So the smallest positive value of ⲫ equivalent to 7.5π will be = 7.5π-2nπ by putting n = 3 i.e., 7.5π-6π 

= 1.5π

y = A.sin2π(t/T-x/λ) ....... (i) 

yₘₐₓ = A. {Because maximum value of sine = 1} 

v = ∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T 

→vₘₐₓ = 2πA/T. {} Because also maximum value of cosine =1} 

a = ∂²y/∂t² = A.sin2π(t/T-x/λ)*4π²/T² 

→aₘₐₓ = 4π²A/T² 

Now yₘₐₓ/vₘₐₓ =A/(2πA/T) = T/2π and vₘₐₓ/aₘₐₓ = (2πA/T)/(4π²A/T²) =T/2π 

Hence, yₘₐₓ/vₘₐₓ = vₘₐₓ/aₘₐₓ.           

We cannot use Componendo and Dividendo because addition and subtraction are allowed only between quantities having same unit or dimension. Here yₘₐₓ, vₘₐₓ and aₘₐₓ all have different units and dimensions.

(a) V = f λ

330 = f × 66 

f = 330/60 = 5.5 Hz

(b) Infrasonic (∠ 20Hz)

Answer is (a) Distance between A and C

V = 340 m/s 

λ = 0.01 m 

v = u λ 

f = V/λ 

f = 340/0.01 

= 34000 Hz > 20,000 Hz 

Human beings cannot hear these sounds. 

Audible frequency range is 20 Hz to 20000 Hz

(a) Loudness decreases 

(b) Loudness increases

The maximum displacement from mean position called Amplitude.

The wave equation can also be written as y = A.sin2π(t/T-x/λ) 

where y is the transverse displacement of the particle at a distance x at time t and A is the amplitude of the particle. The speed of the particle can be found out by partial differentiation of y with respect to t.

 ∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T 

→V = {2πA.cos 2π(t/T-x/λ)}/T 

For V to be maximum cos2π(t/T-x/λ) = 1 

So Vₘₐₓ = 2πA/T

 if A =λ/2π 

Vₘₐₓ = λ/T = v, where v is the wave speed. 

But if A < λ/2π, then Vₘₐₓ < λ/T 

→Vₘₐₓ < v. 

So, if the amplitude is less than the wavelength divided by 2π, the particle speed can never be more than the wave speed.

The wave moves forward or horizontally.

1. The amplitude = 1. 5 m

2. Speed =  \(\frac{Distance\,travellaed}{time}\)

3. V = 800/2 = 400 m/s

λ = 4m 

v = 400 m/s 

f = V/ λ = 400/4 = 100 Hz

In closed organ pipe.

(i) The elevated portions are called crests. The depressed portions are called troughs

(ii) A, C, E, G, I, K, M

(iii) 4 crests, and 3 troughs.

(iv) No

(v) E, I, M

(vi) G, K

(vii) Wavelength = 4 m

Characteristics of Waves:

1. Amplitude : Amplitude is the maximum displacement of a particle from its mean position. This is denoted by the letter a.

2. Wavelength: Wavelength is the distance between two consecutive particles which are in the same phase of vibration. This is equivalent to the distance advanced by the wave by the time a particle has completed one vibration. The Greek letter X (lambda) is used to denote the wavelength. The unit is metre (m).

3. Frequency: Frequency is the number of vibrations in one second.

Frequency = \(\frac{number\,of\,vibrations}{Time\,taken}f=\frac{n}{t}\)

The unit of frequency is hertz (Hz).

(viii) Frequency of the wave = \(\frac{number\,of\,vibrations}{Time}\)

\(f=n/t=\frac{100}{5}=20\)Hz

The equation connecting velocity, wavelength, and frequency of a wave is v = f λ ;

v – Velocity (distance travelled by the wave in one second);

f – frequency (number of vibrations in one second);

x – wavelength (distance between two consecutive particles which are in the same phase of vibration).

(a) 2 cm

(b) 8 m

(c) Frequency, f = \(\frac{n}{t}\)

\(=\frac{3}{0.2}=\frac{30}{2}=15\) Hz

(d) V = fλ = 15 x 8 = 120 m/s

(i)

• wavelength of the first wave = 4m
• wavelength of the second one = 2m

(ii) First wave has a higher wavelength.

(iii) Frequency of the first wave

f = n/t

= \(\frac{3}{0.25}=\frac{300}{25}=12\) Hz

frequency of the second wave

f = \(\frac{6}{0.25}=\frac{600}{25}=24\) Hz

(iv) As frequency increases, wavelength decreases. Wavelength of a wave with a constant speed decreases with increase in frequency, ie. frequency is inversely proportional to the wave length.

f ∝ 1/λ

Pitch is more for shrill sound.

There are two types of mechanical waves: 

They are

1. Transverse waves

2. Longitudinal waves

1. Mechanical waves: The presence of a medium is required for the transmission of these waves, eg. The waves formed on the surface of water, sound waves etc.

2. Electromagnetic waves: Electromagnetic wave is a combined form of an electric field and a magnetic field which vary continuously. A medium is not essential for its propagation eg. radio waves, light waves.

• waves on water
• waves formed in a slinky
• waves formed when a rope is moved up and down after tying one of its ends.
• waves formed on stretched strings.

When we clap our hands from a field or valley having width greater the 17m, we can hear echo. When we talk loudly standing in an auditorium having length more than 17m, we can hear echo.

In wave motion, energy is transferred through the motion of particles of the medium.

The motion in a medium in which, when particles are displaced, a force proportional to the displacement acts on the particles to restore them to their original position is called an elastic wave. If a material has the property of elasticity and the particles in the medium are set into vibratory motion, an elastic wave will be propagated. For example, gas is an elastic medium and sound is transmitted through a gas as an elastic wave.

Wave Propagation Constant (k) : During vibration the phase difference between the particles which are situated at a unit distance is called the wave propagation constant. It is also called the angular wave number. Since the phase difference is 2π in the particles situated at λ distance. Hence for unit distance;

\(k = \frac{2 \pi}{\lambda}\)

For the effective propagation of the waves the medium should have the following properties :

1. The medium should have the elasticity property so that if a particle is displaced then it comes back to the original (initial) position and displaces the particle close to it.

2. For efficient transfer of energy the material medium should have the property of inertia so that it can collect, the energy otherwise the energy would be lost.

3. Every medium resists the disturbance of the motion; which is called resistance. Due to this the energy of oscillation of the particles reduces. If this resistance is more then the dimension of the disturbance becomes zero which means the propagation of energy is for a very small distance. Hence, the resistance of the medium should be very small.

Wave with higher frequency is shown in graph B.

As both travel through the same medium, velocity remains the same. But wavelength is inversely proportional to the frequency. ‘B’ has lower wavelength, so B itself has higher frequency.

When a wave is reflected from a denser medium, its phase changes by π.

The walls are made rough to avoid the regular reflection of sound. Rough surfaces can absorb sound.

If the distance between the walls in a room is more than 17m, the sound waves are reflected repeatedly from the walls, ceiling and floor of the hall, and produce many echos So the sound becomes blurred, distorted and confusing due to overlapping of different sounds.

The methods to minimize the problems that occur due to reflection of sound are

• Use curtains having many folds. 
• Provide large number of ventilators and windows. 
• Make the walls and roof rough. 
• Make sure that the ratio between the height and width of the hall 2:3. 
• Make the floor rough using carpets. 
• Increase the number of audiences.

Whispering galley:

The Whispering Gallery at St. Paul’s Cathedral in London is the best example for the reflection of, sound. Event if you are only whispering near the circular wall below the dome the sound will be heard loudly anywhere within the gallery. This is due to the multiple reflection of sound from the circular walls. The Gol Gumbas in Bijapur of Karnataka is another example.

(i) Ultrasonic waves that are reflected back after striking the object reach the detector. The detector converts the ultrasonic waves into electrical signals.

(ii) Distance = speed × time

(iii) Distance = speed × time .

= 1522 x \(\frac{0.5}{5}\) = 380.5 m

(iv) Bats produce ultrasonic sounds and the sound gets reflected after striking the prey. It can receive the waves.

2d = v × t = 340 × 1 = 340

∴ d = 340/2 = 170m

Distance travelled by the sound

= 2 x 99 = 198 m

time = 0.65

Velocity = \(\frac{Distance}{time}\)

V = 198/0.6

= 330 m/s

Intensity of a wave
I = 2π² n² α² ρ.υ
∴ The intensity of wave depends on :
(i) Frequency; I ∝ n²
(ii) Amplitude I ∝ α²
(iii) Density of a medium 7 ∝ ρ
(iv) Velocity of a wave I ∝ υ

Limitations of Doppler’s Effect:

To observe (view) Doppler’s effect it is necessary that the source, the observer and the medium the velocity of all should be less than the velocity of sound. If their velocity is more than the velocity of sound then the wave velocity graph is distorted and shocking waves are generated. Its example is Jet plane, when the speed of jet plane is more than the speed of sound then Doppler’s effect is not seen.

In incidents related to sound, beats are very important and has many uses. For tuning of musical instruments, this is used. If we wish to listen only one type of note from the musical instrument then it is only possible if it does not have beats. When beats are not present the musical instrument become same tone. With the help of this the frequency of unknown tuning fork is also known.

By the method of beats the calculation of the frequency of unknown tuning fork.

To calculate the frequency of unknown tuning fork, an approximately same but known frequency tuning fork is taken because to observe the beats of sound waves is necessary that the difference in the frequency of both the forks is not much otherwise beats will be formed so fast that to hear them will not be possible. Both the tuning forks are vibrated together and per second the number of beats is calculated. Suppose known frequency of the tuning fork is n and the obtained beats is ∆n then the frequency of unknown tuning would be n + ∆n or n – ∆n.

We know that vibrations in a stretched string is an example of transverse standing waves. To understand it we study a string which is tied between two stationary points and is stretched. If T is the tension in the string and mass of unit length is m, then the velocity of waves produced in the string from equation.

\(v = \sqrt{\frac{T}{m}}\)

When this string is stretched in the perpendicular direction slightly and then left. Then the transverse progressive waves start moving. These waves are reflected from the rigid ends of the string. Therefore, by superposition of incident and reflected waves transverse standing waves are generated and these waves remain till their energy is not destroyed due to friction, etc. Since both the ends of the string are tied up hence they are always nodes.

When the string is stretched from middle and is left then the string vibrates in one part. Both the ends of the string are nodes and the middle point is antinode figure. This is the simple method of producing standing wave in a stretched string in which the string vibrates in a loop.

Let A and B be two tuning forks, then frequency of A
n_A = 256 Hz
Tuning fork b produces 6 beats with fork B.
∴ n_B = n_A ± n = 256 ± 6 = 250 or 262 Hz
∵Fork B produces 3 beat/s with the other fork (frequency 253 Hz), it is only possible when.
n₃ = 250 Hz

f = 20 Hz, 

v = f λ , 

λ = v/f = 17m 

f = 20000Hz, 

v = f λ, 

λ = 0.017 m 

so limit of wavelength = 0.017 to 17 

= 0.017 to 17 m

Transverse wave: a, d, f 

Longitudinal wave: b, c, e