## NCERT Solutions Class 11 Physics Chapter 15- WavesNCERT Solutions for Class 11 Physics Chapter 15 Waves. The subject-matter experts at Javatpoint developed Waves with the intention of assisting students in their exam preparation. Students can learn more about the chapter and get a quick review before their exam by practicing questions from the NCERT textbook using the solutions. NCERT Solutions for Class 11 Physics offer a solid foundation of fundamental concepts that will aid students in their higher levels of education. These solutions were developed in accordance with the most recent CBSE Syllabus 2022-23. A wave is essentially a quivering disturbance that happens when particles move repeatedly through a medium. In Chapter 15 of the NCERT Solutions, students learn how to lay a solid foundation for their future studies in disciplines like engineering and medical science. In this chapter, new ideas including the Doppler effect and the interactions between various wave types will be covered with the students. It will give pupils a comprehensive understanding of the crucial subjects, assisting them in achieving high scores on the Class 11 exam. ## NCERT Solutions Class 11 Physics Chapter 15
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Mass of the string, M = 2.50 kg Tension in the string, T = 200 N Length of the string, l = 20.0 m Therefore, the transverse wave take 40 m/s to reach other side.
A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top, given that the speed of sound in air is 340 m s
Height of the bridge, s = 300 m The initial velocity of the stone, u = 0 Acceleration, a = g = 9.8 m/s Speed of sound in air = 340 m/s Let us consider that the stone hit the water surface after time interval t. Thus, from equation of motion we can write that,
Steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s
Length of the steel wire, l = 12 m Mass of the steel wire, m = 2.0 kg Velocity of the transverse wave, v = 343 m/s
Using the formula, Explain why the speed of sound in air: (a) does not depend upon pressure. (b) increases with temperature and humidity. (c) increases with humidity.
We are given the relation, From ideal gas equation, we know that, a.) When the temperature of the gas is constant It means that the velocity of the sound does depend upon the pressure of t he gas. b.) Since, So speed of the sound increases with temperature. c.) The effective density of the air decreases as humidity rises. This implies that: thus velocity increases.
We know that the function y = f (x, t) represents a wave travelling in one direction, where x and t must appear in the combination x + vt or x - vt or, i.e. y = f (x ± vt). Is the converse true? Can the following functions for y possibly represent a travelling wave:
No, the converse is not true because it is necessary for a wave function representing a travelling wave to have a finite value for all values of x and t. As none of the above functions satisfies the given condition, none of the options represents a travelling wave.
A bat emits the ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound and (b) the transmitted sound? The speed of sound in air is 340 m/s, and in water, 1486 m/s.
Frequency of the ultrasonic sound, v = 1000 kHz = 10 Speed of the sound in air, v a.) the reflected sound b.) the transmitted sound
An ultrasonic scanner operating at 4.2 MHz is used to locate tumors in tissues. If the speed of sound is 2 km/s in a certain tissue, calculate the wavelength of sound in this tissue.
Speed of sound in the tissue,v Operating frequency of the scanner,ν = 4.2 MHz = 4.2 × 10 Thus, the wavelength of the sound is given by,
A transverse harmonic wave on a wire is expressed as: - Is it a stationary wave or a travelling one?
- If it is a travelling wave, give the speed and direction of its propagation.
- Find its frequency and amplitude.
- Give the initial phase at the origin.
- Calculate the smallest distance between two adjacent crests in the wave.
[X and y are in cm and t in seconds. Assume the left to the right direction as the positive direction of x]
The given equation is, i.) Equation of a progressive wave travelling from right to left is given by, On comparing equation (1) with equation (2). we can see that it represents a wave travelling from right to left. Therefore, ii.) Thus, speed of propagation, iii.) Amplitude of the wave, a = 3 cm iv.) Initial phase at the origin = π/4 v.) Smalles distance between two adjacent crests in wave is,
For the wave in the above question (Q8), plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. (i) Give the shapes of these plots. (ii) With respect to which aspects (amplitude, frequency or phase) does the oscillatory motion in a travelling wave differ from one point to another?
We are given the equation: Plotting the displacement (y) vs (t) graphs using different values of t listed below: (ii) Similarly, graphs are obtained for x = 0, x = 2 cm, and x = 4 cm. Only the phase distinguishes the oscillatory motion in the travelling wave from one another. For any change in x, amplitude and frequency remain constant.
A travelling harmonic wave is given as: y(x,t) = 2.0 cos 2π (10t - 0.00580x + 0.35) What is the phase difference between the oscillatory motion of two points separated by a distance of: - 8 m
- 1 m
- λ /2
- 6λ/4
[X and y are in cm, and t is in secs].
The equation for a travelling harmonic wave: y(x,t) = 2.0 cos 2π (10t - 0.00580x + 0.35) = 2.0 cos (20πt - 0.016πx + 0.70π) Where, Propagation constant, k = 0.0160 π Amplitude, a = 2 cm Angular frequency, ω = 20 π rad/s
The transverse displacement of a wire (clamped at both its ends) is described as: The mass of the wire is 6 x 10 Provide answers to the following questions: (i) Is the function describing a stationary wave or a travelling wave? (ii) Interpret the wave as a superposition of two waves travelling in opposite directions. Find the speed, wavelength and frequency of each wave. (iii) Calculate the wire's tension. [X and y are in meters and t in secs]
The standard equation of a stationary wave is described as: y(x,t) = 2a sin (kx) cos (ωt) The given equation is: The given above equation is similar to the general equation. i.) Hence we can conclude that the given function describes a kind of stationary wave. ii.) When a wave is travelling along the positive direction of x - axis then it can be represented as: y Also, when a wave is travelling along the negative direction of x - axis then it can be represented as: y If we super-position these two waves, then it gives the following result, iii.) Wire's tension. The velocity of the transverse wave,v = 180 m/s The string's mass,m = 6 ×10 String length,l = 3 m
Considering the wave described in the above question (Q11), answer the following questions: (a) Are all the points in the wire oscillating at the same values of - frequency,
- phase, and
- amplitude?
Justify your answers. (b) Calculate the amplitude of a point 0.4 m away from one end.
a.) Since the wire is clamped at both ends, the ends act as nodes and cause a single segment of vibration to run through the entire wire. Thus, (i) Every particle in the wire oscillates at the same frequency, with the exception of the ends. This is because the nodes which are present at the ends are having frequency equal to zero. (ii) Since every particle in the wire is contained within a single segment, every particle has the same phase. Except for the nodes. (iii) However, amplitude varies for various points. b.) The given equation is,
Given below are functions of x and t to describe the displacement (longitudinal or transverse) of an elastic wave. Identify the ones describing (a) a stationary wave, (b) a travelling wave and (c) neither of the two: (i) y = 3 sin( 5x - 0.5t ) + 4cos( 5x - 0.5t ) (ii) y = cosx sint + cos2x sin2t. (iii) y = 2 cos (3x) sin (10t) (iv)
(i) The given equation is, y = 3 sin (5x - 0.5t) + 4 cos(5x - 0.5t) In the above equation we can see that the harmonic terms ωt and kx are in the combination form form of kx - ωt. Thus we can conclude that the following equation describes a travelling wave. ii.) The given equation is, y = cos (x) sin (t) + cos (2x) sin (2t) In the above equation the harmonic terms ωt and kx appears separately therefore it represents a stationary wave. Further, the given equation describes the superposition of two stationary waves. iii.) The given equation is, y = 2 cos (3x) sin (10t) In the above equation the harmonic terms ωt and kx appears separately therefore we can conclude that the given equation represents a stationary wave. iv.) The given equation is, As we can see that there is no harmonic term in this equation. As a result, it is neither a wave that travels nor one that is fixed.
A string clamped at both ends is stretched out, it is then made to vibrate in its fundamental mode at a frequency of 45 Hz. The linear mass density of the string is 4.0 × 10 Calculate: - The velocity of a transverse wave on the string
- The tension in the string
Mass of the string, m = 2 × 10 Linear density of the string = 4 × 10 Frequency of the string,v We know the length of the wire is given by the relation,
A 1 m long pipe with a movable piston at one end and an opening at the other will be in resonance with a tuning fork vibrating at 340 Hz, if the length of the pipe is 79.3 cm or 25.5 cm. Calculate the speed of sound in the air. Neglect the edge effects.
Frequency of the turning fork, v Length of the pipe, l The supplied pipe will behave as a pipe with one end closed and the other open since the pipe has a piston at one end, as shown in the following figure: These systems produce strange harmonics. We know that a basic note in a closed pipe can be written as follows:
A steel bar of length 200 cm is nailed at its midpoint. The fundamental frequency of the longitudinal vibrations of the rod is 2.53 kHz. At what speed will the sound be able to travel through steel?
Length of the steel bar,l = 200 cm = 2 m Fundamental frequency of vibration,v Antinodes (A) and nodes (N) are then formed at the bar's core and its two edges, as shown in the following diagram:
One end of A 20 cm long tube is closed. Find the harmonic mode of the tube that will be resonantly excited by a source of frequency 430 Hz. If both ends are open, can the same source still produce resonance in the tube? (Sound travels in air at 340 m /s).
Length of the pipe,l = 20 cm = 0.2 m Frequency of the source = n The normal mode of frequency,v Speed of sound,v = 340 m/s We know that in a closed pipe, the nth normal mode of frequency is given by, As a result, the first mode of vibration frequency is excited resonantly by the provided source. Now, for a pipe open at both ends, the nth mode of vibration frequency: This source does not vibrate in resonance with the tube because the mode of vibration (n) must be an integer.
Guitar strings X and Y striking the note 'Ga' are a little out of tune and give beats at 6 Hz. When the string X is slightly loosened, and the beat frequency becomes 3 Hz. Given that the original frequency of X is 324 Hz, find the frequency of Y.
Frequency of X, f Frequency of Y = f Beat's frequency,n = 6 Hz Also,n = |f ∴ 6 = 324 ± f f As frequency drops with a decrease in tension in the string, thus f
Explain how: (i) A sound wave's pressure antinode is a displacement node and vice versa. (ii) The Ganges river dolphin, despite being blind, can manoeuvre and swim around obstacles and hunt down prey. (iii) A guitar note and violin note are being played at the same frequency, however, we can still make out which instrument is producing which note. (iv) Both transverse and longitudinal waves can propagate through solids, but only longitudinal waves can move through gases. (v) In a dispersive medium, the shape of a pulse propagating through it gets distorted.
(i) An antinode is a location where pressure is at its lowest and vibration amplitude is at its highest. A node, on the other hand, is a location where vibration amplitude and pressure are both at their lowest points. (ii) The Ganges river dolphin uses click noises to communicate with itself about the positions and distances of objects in front of it. The click noises are returned to the dolphin as vibrations. As a result, it can move around and hunt its prey with little vision. (iii) The overtones produced by the violin and guitar are of varying intensities. As a result, even though notes from a guitar and a violin vibrate at similar frequencies, they may be distinguished. (iv) The bulk modulus of elasticity exists in both solids and liquids. As a result, they both permit the passage of longitudinal waves through them. Gases do not, however, have a shear modulus like solids do. Transverse waves cannot therefore travel through gases. (v) A pulse is made up of several waves with distinct wavelengths. In a dispersive medium, these waves flow at various speeds. Its shape is distorted as a result.
A train, standing at the outer signal of a railway station, blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m/s, (b) recedes from the platform with a speed of 10 m/s? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m/s.
Frequency of the whistle = 400 Hz Speed of sound in still air = 340 m/s i.) Frequency of the whistle for a platform observer (ii) The speed of the sound will not change. It will remain at 340 m/s.
A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m/s. What are the frequency, wavelength, and speed of sound for an observer standing on the station's platform? Is the situation exactly identical to the case when the air is still, and the observer runs towards the yard at a speed of 10 m/s? The speed of sound in still air can be taken as 340 m/s.
Frequency of the whistle = 400 Hz Speed of wind,v Speed of sound in still air,v = 340 m/s The effective speed of the sound for an observer standing on the platform is: v = 340 + 10 = 350 m/s Since the source and the observer are not moving relative to one another, the observer will hear the same frequency of sound. Therefore, frequency,f = 400 Hz Wavelength of the sound heard by the observer, There is a relative motion between the observer and the source with regard to the medium when the air is static and the observer moves towards the garden at a speed of 10 m/s. The medium is at rest. Therefore, Thus, we can see that the situations in the two cases are entirely different.
A travelling harmonic wave on a string is described by, (a) What are the displacement and velocity of oscillation of a point at x = 1 cm and t = 1 s? Is this velocity equal to the velocity of wave propagation? (b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
a.) The travelling harmonic wave is given as: The velocity of oscillation, At x = 1 cm and t = 1 s The standard equation is given as: The velocity of the wave propagation, Velocity at x = 1 cm and t = 1 sec is not equal to the velocity of wave propagation. b.) Propagation constant, All the points at a distance ± λ, ± 2λ,...from x = 1 cm will have the same transverse displacement and velocity. As λ = 12.56 m, the points ± 12.56m, ± 25.12m,... and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2s,5s and 11s.
A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, and (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s (that is, the whistle is blown for a split of seconds after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
(a) The propagation speed is fixed; it is equal to the sound speed in air. The frequency and wavelength won't be known with certainty. (b) A whistle's note does not have a frequency of 1/20 = 0.05 Hz. However, the whistle's short pip repeats at a frequency of 0.05 Hz.
One end of a long string of linear mass density 8.0 × 10
Linear mass density of the string, μ = 8.0 × 10 Frequency of the tuning fork = 256 Hz Mass on the pan = 90 kg Tension on the string,T = 90 × 9.8 = 882 N Amplitude, A = 0.05 m For a transverse wave, the velocity is given by the relation.
A SONAR system fixed in a submarine operates at a frequency of 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km/h. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m/s.
Frequency of the SONAR system,f = 40 kHz = 40 × 10 Speed of sound in water,v = 1450m/s Speed of the enemy submarine,v The energy submarine approaches the SONAR, which is at rest. Therefore, the relation provides the apparent frequency.
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about 4.0 km/s, and that of the P wave is 8.0 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in a straight line, at what distance does the earthquake occur?
Let the speeds of S and P be v
A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in the air. What frequency does the bat hear reflected off the wall?
The sound emission frequency of the bat = 40 kHz The velocity of the bat, v Where v is the velocity of the sound in the air. The apparent frequency of the sound hitting the wall: The frequency gets reflected by the wall and is received by the bat moving towards the wall:
A man standing at a certain distance from an observer blows a horn of frequency 200 Hz in still air. (a) Find the horn's frequency for the observer when the man (i) runs towards him at 20 m/s (ii) runs away from him at 20 m /s. (b) Find the speed of sound in both cases. [Speed of sound in still air is 340 m/ s]
Frequency of the horn,v Velocity of the man,v Velocity of the sound,v = 340 m/s a.) Horn's frequency, i.) When the man runs towards observer The apparent frequency of the horn as the man approaches the observer is: ii.) When the man runs away from the observer The apparent frequency of the horn as the man runs away from the observer is: (b) The speed of sound is 340 m/s in both cases. The apparent change in frequency is a result of the relative motions of the observer and the source.
A truck parked outside a petrol pump blows a horn of frequency 200 Hz in still air. The wind then starts blowing towards the petrol pump at 20 m/s. Calculate the wavelength, speed, and frequency of the horn's sound for a man standing at the petrol pump. Is this situation completely identical to a situation when the observer moves towards the truck at 20 m/s and the air is still?
Let us consider the case when the man is standing at the petrol pump: Frequency is,v Velocity of the sound,v = 340 m/s Speed of the wind,v The observer will perceive the sound at 200 Hz because there is no relative motion between them (the observer standing and the truck). Wind is blowing in observer's direction at a speed of = 20 m/s Effective velocity of the sound,v The wavelength ( λ ) of the sound: Now, consider the case when the observer is running towards the train: Speed of the observer,v The apparent frequency of the sound as the observers move towards the truck is: Next TopicNext Topic# |