DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

DC Pandey Solutions for JEE Physics

JEE : DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

The document DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

Introductory Exercise 15.1

Q.1. A string vibrates according to the equation DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev where x and y are in centimetres and t is in seconds.
(a) What is the speed of the component wave?
(b) What is the distance between the adjacent nodes?
(c) What is the velocity of the particle of the string at the position x = 1.5 cm when t = 9/8 s?

Sol. 

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
(b) Distance between adjacent nodes
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
Now substitute the given values of x and t.

Q.2. If two waves differ only in amplitude and are propagated in opposite directions through a medium, will they produce standing waves? Is energy transported?
Sol.
In this case, node points will also oscillate, but standing waves are formed.

Ques 3: Figure shows different standing wave patterns on a string of linear mass density 4.0 × 10-2 kg/m under a tension of 100 N. The amplitude of antinodes is indicated in each figure. The length of the string is 2.0 m.
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
(i) Obtain the frequencies of the modes shown in figures (a) and (b). 

(ii) Write down the transverse displacement y as a function of x and f for each mode. (Take the initial configuration of the wire in each mode to be as shown by the dark lines in the figure).
Sol: 

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

Q.4. A 160 g rope 4 m long is fixed at one end and tied to a light string of the same length at the other end. Its tension is 400 N.
(a) What are the wavelengths of the fundamental and the first two overtones? 
(b) What are the frequencies of these standing waves?
[Hint: In this case, the fixed end is a node and the end tied with the light string is antinode.]
Sol.

(a) 1 = λ1/4
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
Now DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
Similarly f2 and f3.

Q.5. A string fastened at both ends has successive resonances with wavelengths of 0.54m for the nth harmonic and 0.48 m for the (n +1)th harmonic.
(a) Which harmonics are these?
(b) What is the length of the string?
(c) What is the wavelength of the fundamental frequency?
Sol.

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
or
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev ...(i)
Similarly,
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
or
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev ... (ii)
From Eqs. (i) and (ii) we have,

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
(b) From Eq. (i)
l = (0.27)n = 0.27 × 8 = 2.16 m
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

Q.6. If the frequencies of the second and fifth harmonics of a string differ by 54 Hz. What is the fundamental frequency of the string?
Sol.

5f - 2f=54
3f = 54 f = 18 Hz

Q.7. A wire is attached to a pan of mass 200 g that contains a 2.0 kg mass, as shown in the figure. When clucked, the wire vibrates at a fundamental frequency of 220 Hz. An additional unknown mass M is then added to the pan and a fundamental frequency of 260 Hz is detected. What is the value of M1

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
Sol.

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
Solving we get,
M = 0.873 kg

Q.8. A wire fixed at both ends is 1.0 m long and has a mass of 36 g. One of its oscillation frequencies is 250 Hz and the next higher one is 300 Hz.
(a) Which harmonics do these frequencies represent?
(b) What is the tension in the wire?
Sol.

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev
So, f1 is 5th harmo nic and f2 is 6th harmonic.
DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Extra Questions

,

ppt

,

Exam

,

Important questions

,

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

,

Previous Year Questions with Solutions

,

Free

,

Objective type Questions

,

MCQs

,

Viva Questions

,

Semester Notes

,

video lectures

,

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

,

practice quizzes

,

study material

,

past year papers

,

DC Pandey Solutions: Superposition of Waves- 1 JEE Notes | EduRev

,

pdf

,

mock tests for examination

,

Sample Paper

,

shortcuts and tricks

,

Summary

;