NCERT Solution (Class 11) > Physics Chapter 15 Waves
NCERT Solution Class 11 Physics Chapter 15 Waves: National Council of Educational Research and Training Class 11 Physics Chapter 15 Solution – Waves. Free PDF Download facility available at our website.
Board |
NCERT |
Class |
11 |
Subject |
Physics |
Chapter |
15 |
Chapter Name |
Waves |
Topic |
Exercise Solution |
Waves Chapter all Questions and Numericals Solution
Chapter 15 – Waves
15.3)
Answer-
15.5) You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave :
(a) (x – vt ) 2
(b) log [(x + vt)/x0 ]
(c) 1/(x + vt)
ANSWER-
Given function is showing infinite value mean not finite value hence it is travelling wave.
15.6)
15.7)
15.8) A transverse harmonic wave on a string is described by y (x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation? (b) What are its amplitude and frequency? (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?
Answer-
A transverse harmonic wave on a string is described by
15.9) For the wave described in Exercise 15.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?
Answer-
A transverse harmonic wave on a string is described by
y(x, t) = 3.0 sin (36 t + 0.018 x + π/4)
for x = 0 ,
∴ y(x, t) = 3.0 sin (36 t + π/4)
For different values of t we can obtain values of y
for x = 2 cm
∴ y(x, t) = 3.0 sin (36 t +0.036 + π/4)
From above formula For different values of t we can obtain values of y
for x = 4 cm
∴ y(x, t) = 3.0 sin (36 t +0.072 + π/4)
From above formula For different values of t we can obtain values of y
∴ The y-t plots of the three waves shown below;
15.10)
15.11)
Answer –
15.12) (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
ANSWER-
Given wave is stationary wave hence at point where string is clamped is called node and which is fixed hence have zero amplitude ,zero frequency and zero phase angle while all other points oscillates with same frequency and same phase angle but different magnitudes.
15.14)
15.15)
In case you are missed :- NCERT Solution for Oscillations
15.16)
15.17)
15.18)
15.19)
Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”, (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium.
Answer:
(a) A node is a point where the amplitude of vibration is the minimum and pressure is the maximum. While an antinode is a point where the amplitude of vibration is the maximum and pressure is the minimum. Therefore, a displacement node is a pressure antinode, and vice versa.
(b) Bats emit ultrasonic sound waves. These ultrasonic sound waves travel and reflected back from obstacle towards bat. A bat receives a reflected wave and calculates the distance, direction, nature, and size of an obstacle without any eyes but by using brain senses.
(c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes because the overtones and strength produced by a sitar and a violin, are different.
(d) Solids have definite shape hence they possess the shear modulus. On the other hand fluids do not have any definite shape. The propagation of a transverse wave is such that it produces shearing stress in a medium hence The propagation of such a wave is possible only in solids, and not in gases.
Both solids and fluids have bulk modulus property as they have volume. Longitudinal waves travels with compression and rare fraction so both solid and gases withstand the compression due to property of bulk modulus hence. Longitudinal waves can propagate through solids as well as fluids.
(e) Dispersive medium is the medium in which waves travels with different velocities depends upon nature of medium. A pulse is a combination of waves. These waves travel in a dispersive medium with different velocities hence distortion of the shape of a wave pulse occurs.
15.20)
Additional Exercises
15.22)
15.23)
15.26)
15.27)
In case you are missed :- NCERT Solution for Kinetic Theory