NCERT Solutions for Class 11 Physics Chapter 15 Waves

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.

Answer: No, the converse is not true. The basic requirement for a wave function to represent a travelling wave is that for all values of x and t, wave function must have a finite value. Out of the given functions for y none satisfies this condition. Therefore, none can represent a travelling wave.

Physics Waves Notes and NCERT Solution of Class 11th.

Question 15. 7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 km s-1? The operating frequency of the scanner is 4.2 MHz.
Answer: Here speed of sound => υ = 1.7 km s-1 = 1700 ms-1 and
frequency υ= 4.2 MHz = 4.2 x 106 Hz
.’. Wavelength, A = υ/V = 1700/(4.2 x 106) =4.1 x 10-4 m.

Question 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:

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.

Question 15. 10. For the travelling harmonic wave
y(x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)
where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of
(a) 4 m (b) 0.5 m
(c) λ/2 (d) 3λ/4.
Answer:

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.

Number of beats, b = 6
υ2 = υ1 ± b = 324 ± 6 !.e., υ2 = 330 Hz or 318 Hz
Since the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency υ1 will be reduced i.e., number of beats will increase if υ2 = 330 Hz. This is not so because number of beats become 3.
Therefore, it is concluded that the frequency υ2 = 318 Hz. because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with υ2 = 318 Hz.

Question 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) In a sound wave, a decrease in displacement i.e., displacement node causes an increase in the pressure there i.e., a pressure antinode is formed. Also, an increase in displacement is due to the decrease in pressure.
(b) Bats emit ultrasonic waves of high frequency from their mouths. These waves after being reflected back from the obstacles on their path are observed by the bats. These waves give them an idea of distance, direction, nature and size of the obstacles.
(c) The quality of a violin note is different from the quality of sitar. Therefore, they emit different harmonics which can be observed by human ear and used to differentiate between the two notes.
(d) This is due to the fact that gases have only the bulk modulus of elasticity whereas solids have both, the shear modulus as well as the bulk modulus of elasticity.
(e) A pulse of sound consists of a combination of waves of different wavelength. In a dispersive medium, these waves travel with different velocities giving rise to the distortion in the wave.

Question 15. 20. 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 ms~1. (b) recedes from the platform with a speed of 10 ms-1(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 ms-1   .
Answer: Frequency of whistle, v = 400 Hz; speed of sound, υ= 340 ms-1 speed of train, υs= 10 ms1
(i) (a) When the train approaches the platform (i.e., the observer at rest),

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.

Question 15. 23. 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, (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 second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz?
Answer:  (a) In a non dispersive medium, the wave propagates with definite speed but its wavelength of frequency is not definite.
(b) No, the frequency of the note is not 1/20 or 0.50 Hz. 0.005 Hz is only the frequency ‘ of repetition of the pip of the whistle.

Question 15. 24. One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
Answer: Here, mass per unit length, g = linear mass density = 8 x 10-3 kg m-1;
Tension in the string, T = 90 kg = 90 x 9.8 N= 882 N;

Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes and NCERT Solution of Class 11th.
Physics Waves Notes

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