NIOS Class 12 Physics Chapter 14 Wave Phenomena

NIOS Class 12 Physics Chapter 14 Wave Phenomena Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 14 Wave Phenomena Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.

NIOS Class 12 Physics Chapter 14 Wave Phenomena

Also, you can read the NIOS book online in these sections Solutions by Expert Teachers as per National Institute of Open Schooling (NIOS) Book guidelines. These solutions are part of NIOS All Subject Solutions. Here we have given NIOS Class 12 Physics Notes, NIOS Senior Secondary Course Physics Solutions in English for All Chapter, You can practice these here.

Wave Phenomena

Chapter: 14

INTEXT QUESTIONS 14.1

1. State the differences between longitudinal and transverse waves?

Ans:

2. Write the relation between phase difference and path difference.

Ans: Phase difference () and path difference (p) are related as:

θ = 2π/λ × p

3. Two simple harmonic waves are represented by equations y₁ = a sin (What is the ωt – k_x) and y₂ = a sin [(ωt – k_x + φ] phase difference between these two waves?

Ans: The two given simple harmonic waves are:

y₁= a sin ⁡ (ωt−k_x)y₁ 

= a sin(ωt – k_x)

y₂ = asin⁡[(ωt−k_x)+ϕ]y₂ 

= a \sin[(\omega t – k_x) + \phi]

Here, ωt−k_x\omega t – k_x is the common phase of both waves, and the second wave has an additional angle ϕ\phi inside the sine function.

The phase difference between two waves is the difference in their arguments (angles inside the sine function).

So, the phase difference between y₂y₂ and y₁y₁ is:

Δϕ = ( ωt− kx+ ϕ)−(ωt − kx) = ϕ

INTEXT QUESTIONS 14.2

1. What was the assumption made by Newton in deriving his formula?

Ans: While deriving his formula for the speed of sound, Newton made the assumption that the propagation of sound in air takes place under isothermal conditions. This means he believed that the compressions and rarefactions caused by the sound wave occur slowly enough for heat exchange to happen between the air particles and their surroundings, thereby keeping the temperature constant throughout the process. Under this assumption, he applied Boyle’s law, which relates pressure and volume at constant temperature. However, this assumption turned out to be inaccurate because sound waves travel rapidly, and the changes in pressure and volume happen too quickly for any significant heat exchange. In reality, the process is adiabatic, not isothermal, meaning no heat is exchanged during the propagation of sound. As a result, Newton’s calculation underestimated the actual speed of sound in air. It was later corrected by Laplace, who accounted for the adiabatic nature of the process by including the adiabatic index (γ) in the formula, thus bringing theoretical values closer to experimental observations.

2. What was wrong with Newton’s formula?

Ans: The problem with Newton’s formula for the speed of sound was that it gave a value much lower than the actual speed observed in experiments. This happened because Newton assumed that sound travels through air in an isothermal process, where the temperature remains constant during compressions and rarefactions. He believed that the air had enough time to exchange heat with its surroundings during the propagation of sound. However, this assumption was incorrect. In reality, the process happens so quickly that there is no time for heat exchange, making it an adiabatic process, where temperature changes occur within the air itself. As a result, Newton’s formula underestimated the speed of sound. Later, Laplace corrected this mistake by considering the adiabatic nature of sound propagation, which gave a value that closely matched experimental results.

3. Show that for every 1°C rise in temperature, the velocity of sound in air increases by 0.61 ms⁻¹.

Ans: From Laplace’s formula:

v=331+0.61t

So, for every 1°C rise, velocity increases by 0.61 ms⁻¹.

4. Calculate the temperature at which the velocity in air is (3/2) times the velocity of sound at 7°C.

Ans: Given: v₁ at t₁ = 7°C

v₂ = 3/2 v₂ 

Let v₁ = 331 + 0.61 × 7

= 335.27 ms^-1

Set v₂ = 3/2 × 335. 27

= 502.905 ms^-1

v₂ = 331 + 0.61 T

So, 502.905

= 331 + 0.61 T 0.61 T = 171.905 T = 171.905 / 0.61

= 282 C

5. Write the formula for the velocity of a wave on stretched string?

Ans: v = √F/m

Where:

F: tension in string

m: mass per unit length

6. Let be the wavelength of a wave on a stretched string of mass per unit length m and n be its frequency. Write the relation between n,,F and m? Further if =2l, what would be the relation between n,l,F and m?

Ans:

INTEXT QUESTIONS 14.3

1. What happens when two waves travelling in opposite directions meet?

Ans: When two waves travelling in opposite directions meet, they undergo a phenomenon called interference, which results in the formation of a stationary wave (also known as a standing wave), if the waves have the same frequency and amplitude.

In this process, the two waves superpose, meaning their displacements add algebraically at each point in space. At certain points, the waves always cancel each other out (destructive interference), creating points of no displacement called nodes. At other points, they reinforce each other (constructive interference), creating points of maximum displacement called antinodes.

These nodes and antinodes remain fixed in space, and the wave pattern appears to stand still, even though energy is still moving through the medium. This is commonly seen in stretched strings, air columns, and other systems where waves reflect and interact.

2. What happens when two marbles each of the same mass travelling with the same velocity along the same line meet?

Ans: When two marbles of the same mass, travelling with the same velocity along the same straight line, meet, the outcome depends on the direction in which they are moving:

Case 1: Moving in the Same Direction:

If both marbles are moving in the same direction with the same velocity, they will never collide because neither one is catching up to the other. They simply continue moving side by side or at the same relative positions indefinitely (assuming no external forces or friction).

Case 2: Moving in Opposite Directions:

If the marbles are moving towards each other (i.e., in opposite directions), they will collide head-on. Since they have:

(i) Equal mass, and

(ii) Equal speed.

and if we assume a perfectly elastic collision then after the collision:

(i) They will simply exchange their velocities.

(ii) Each marble will rebound and move in the opposite direction with the same speed as before the collision.

This is because momentum and kinetic energy are both conserved in an elastic collision between equal masses.

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