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The chapter Waves and Oscillations includes the concepts like periodic motion, oscillatory motion, simple harmonic motion and also gives the answer about “what is oscillating motion”. Oscillation is the process of occurring again and again; the variations in any quantity or measure around its equilibrium value through time.

Therefore, we can define the oscillatory motion as the motion in which a body moves to and fro or back and forth repeatedly about a fixed point i.e, mean position in a fixed interval of time. Some of the oscillation examples are the motion of a pendulum of a wall clock and a girl swinging on a swing.

In simple harmonic motion, it tells us about how the energy, time period and equation of motion of a particle can be given. Also, explain how the expression of the time period of a simple pendulum can be given using this concept.

This chapter also explains about “what is a wave in physics” and various wave types like transverse and longitudinal waves. It also includes concepts like the velocity of longitudinal waves, sound waves and how the stationary waves are formed.

Now, let's move on to the important concepts and formulae related to JEE and JEE Main exams along with a few solved examples.

Simple harmonic motion

Total energy in SHM

Simple pendulum

Oscillation in a loaded spring

Velocity of sound in gases

Standing waves in string and normal mode

Beats

Doppler’s effect in sound

A particle with mass $m$ oscillates about the origin on the x-axis. Its potential energy is calculated as $U(x)=k\,x^4$, where $k$ is a positive constant. If the amplitude of the oscillation is $a$, then calculate its time period of the oscillation.

Sol:

Given that,

Potential energy, $U(x)=k\,x^4$

In order to solve this first we have to use the relation between force and potential energy of a particle and after converting it to standard simple harmonic motion we can obtain the time period of the oscillation.

The relation of force between potential energy is,

$F=-\dfrac{\partial U}{\partial x}$

After partially derivative of $U$, we get;

$F=-\dfrac{\partial (k\,x^4)}{\partial x}$

$F=-4kx^3$ ……….(1)

Here we can write force as,

$F=ma=m \dfrac{\text{d}^2x}{\text{d}t^2}$.......(2)

After putting the value of $F$ from equation (2) into equation (1), we get;

$m \dfrac{\text{d}^2x}{\text{d}t^2} =-4kx^3$ …………(3)

The standard equation of SHM is,

$\dfrac{\text{d}^2x}{\text{d}t^2}+\omega^2x=0$

$\dfrac{\text{d}^2x}{\text{d}t^2}=-\omega^2x$.......(4)

Here, $x=a\sin\omega t$

Now putting the value of $\dfrac{\text{d}^2x}{\text{d}t^2}$ from eq.(4) to eq.(3), we get;

$-m \omega^2x=-4kx^3$

Om simplification,

$m\omega^2 = 4kx^2$

$m(\dfrac{2\pi}{T})^2= 4kx^2$.........(As, $\omega=\dfrac{2\pi}{T}$)

$\dfrac{2\pi}{T}=\sqrt{\dfrac{4kx^2}{m}}$

$T=2\pi\sqrt{\dfrac{m}{4kx^2}}$

Hence, the time period of the oscillation is $2\pi\sqrt{\dfrac{m}{4kx^2}}$.

Key Point: The relation between force and potential energy of the particle and the standard equation of simple harmonic motion is crucial for solving such problems.

A sonometer wire of length $1.5\,m$ is made of steel. The tension in it produces an elastic strain $1\%$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^3\,kg/m^3$ and $2.2\times 10^{11}\,N/m^2$ respectively ?

Sol:

Given that,

Length of the wire, $l=1.5\,m$

Elastic strain, $\dfrac{\Delta L}{L}= 1\%$

Density, $\rho= 7.7 \times 10^3\,kg/m^3$

Young’s modulus of elasticity, $Y=2.2\times 10^{11}\,N/m^2$

In order to solve this problem first we have to use the relation of Young’s modulus of elasticity to find the value of force and after that we can put it in the expression of fundamental frequency.

The expression of Young’s modulus is given as,

$Y=\dfrac{F}{A (\Delta L/L)}$

After rearranging we can write;

$F=YA(\dfrac{\Delta L}{L})$

Putting the values of known quantities, we get;

$F=2.2\times 10^{11}\times A \times \dfrac{1}{100}=2.2A \times 10^9$

The expression of fundamental frequency is,

$\nu=\dfrac{1}{2l}\sqrt{\dfrac{T}{m}}$

$\nu=\dfrac{1}{2l}\sqrt{\dfrac{F}{A \times 1 \times \rho}}$

After putting the values of $F$ and $\rho$, we get;

$\nu=\dfrac{1}{2 \times 1.5}\sqrt{\dfrac{2.2A \times 10^9}{A \times 1 \times \rho}}$

Upon simplification, we get;

$\nu=\dfrac{10^3}{3}\sqrt{\dfrac{2}{7}}$

$\nu=178.2\,Hz$

Hence, the fundamental frequency of steel is $178.2\,Hz$.

Key Point: The expression of Young’s modulus and fundamental frequency is essential to solve this problem.

A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to- (JEE Main 2019)

a. 16.6 cm

b. 10.0 cm

c. 20.0 cm

d. 33.3 cm

Sol:

Given that,

Length of the string, $L=1\,m$

Mass of the string, $M=5\times 10^{-3}\,kg$

Hence, mass per unit length, $m=\dfrac{M}{L}=\dfrac{5\times 10^{-3}}{1}\,kg/m$

Tension in the string, $T=8.0\,N$

Frequency, $\nu=100\,Hz$

In order to solve this problem, we first use the relation of speed of the transverse wave in the string and after that using the relation between the frequency and wavelength we are able to find the distance between the successive nodes.

Now the expression for speed of transverse waves in string is given as,

$v=\sqrt{\dfrac{T}{m}}$...........(i)

And the wavelength-frequency relationship is given as,

$\nu=\dfrac{v}{\lambda}$

$v = \nu \lambda$............(ii)

Now, putting the value of $v$ using eq.(ii) in eq.(ii),

$\nu \lambda = \sqrt{\dfrac{T}{m}}$

$\lambda =\dfrac{1}{\nu} \sqrt{\dfrac{T}{m}}$

After putting the values of all quantities, we get;

$\lambda =\dfrac{1}{100} \sqrt{\dfrac{8.0}{5\times 10^{-3}}}$

$\lambda = \dfrac{40}{100} = 0.40\,m=40\,cm$

The distance between the two successive nodes is $\dfrac{\lambda}{2}$, therefore it is $\dfrac{40}{2}\,cm=20\,cm$.

Hence, option c is correct.

Key Point: The relation of speed of transverse wave and frequency of the wave is important to solve such a problem.

A tuning fork vibrates with frequency $256\,Hz$ and gives one best per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound of air is $340\,m/s$) (JEE Main 2018)

a. 190 cm

b. 180 cm

c. 220 cm

d. 200 cm

Sol:

Given that,

Frequency of oscillation of tuning fork =$256\,Hz$

Therefore, the frequency of oscillation of open pipe = $(256\pm 1)\,Hz$

To solve this problem we have to use the relation of the third normal mode of vibration of an open pipe to find the length of the open pipe.

The relation of third normal mode of vibration is given as,

$\nu=\dfrac{3v}{2l}$

After rearrangement, we can write;

$l=\dfrac{3v}{2\nu}$

Noe putting the values of known quantities, we get;

$l=\dfrac{3\times 340}{2\times 255}$

$l= 2.00\,m=200\,cm$

Therefore, the length of the open pipe is $200\,m$. Hence option d is correct.

Key Point: The relation of normal modes of vibration in open pipe and beats per second is important to solve this problem.

1. Two wires are fixed on a sonometer. Their tensions are in the ratio $8:1$, their lengths are in the ratio $36:35$, the diameters are in the ratio $4:1$ and densities are in the ratio $1:2$. If the note of the higher pitch has a frequency $360\,s^{-1}$, then calculate the frequency of beats produced?

(Ans: $10\,s^{-1}$)

2. A particle moves with simple harmonic motion in a straight line. In the first $\tau\,s$, after starting from rest it travels a distance $a$, and in next $\tau\,s$ it travels distance $2a$ in the same direction, then calculates the time period of oscillation.

(Ans: $6\tau$)

In this article, we have discussed the concepts like simple harmonic motion, simple pendulum, standing waves and velocity of sound in gases medium etc. We have also discussed “what is the basic principle of oscillation” and also “what is wave”. Mention various important formulas and concepts of the chapter.

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JEE Main 2022 June and July Session exam dates and revised schedule have been announced by the NTA. JEE Main 2022 June and July Session will now be conducted on 20-June-2022, and the exam registration closes on 5-Apr-2022. You can check the complete schedule on our site. Furthermore, you can check JEE Main 2022 dates for application, admit card, exam, answer key, result, counselling, etc along with other relevant information.

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NTA has announced the JEE Main 2022 June session application form release date on the official website https://jeemain.nta.nic.in/. JEE Main 2022 June and July session Application Form is available on the official website for online registration. Besides JEE Main 2022 June and July session application form release date, learn about the application process, steps to fill the form, how to submit, exam date sheet etc online. Check our website for more details. July Session's details will be updated soon by NTA.

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Download JEE Main Question Papers & Answer Keys of 2021, 2020, 2019, 2018 and 2017 PDFs. JEE Main Question Paper are provided language-wise along with their answer keys. We also offer JEE Main Sample Question Papers with Answer Keys for Physics, Chemistry and Maths solved by our expert teachers on Vedantu. Downloading the JEE Main Sample Question Papers with solutions will help the engineering aspirants to score high marks in the JEE Main examinations.

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In order to prepare for JEE Main 2022, candidates should know the list of important books i.e. RD Sharma Solutions, NCERT Solutions, RS Aggarwal Solutions, HC Verma books and RS Aggarwal Solutions. They will find the high quality readymade solutions of these books on Vedantu. These books will help them in order to prepare well for the JEE Main 2022 exam so that they can grab the top rank in the all India entrance exam.

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JEE Main 2022 free online mock test series for exam preparation are available on the Vedantu website for free download. Practising these mock test papers of Physics, Chemistry and Maths prepared by expert teachers at Vedantu will help you to boost your confidence to face the JEE Main 2022 examination without any worries. The JEE Main test series for Physics, Chemistry and Maths that is based on the latest syllabus of JEE Main and also the Previous Year Question Papers.

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NTA is responsible for the release of the JEE Main 2022 June and July Session cut off score. The qualifying percentile score might remain the same for different categories. According to the latest trends, the expected cut off mark for JEE Main 2022 June and July Session is 50% for general category candidates, 45% for physically challenged candidates, and 40% for candidates from reserved categories. For the general category, JEE Main qualifying marks for 2021 ranged from 87.8992241 for general-category, while for OBC/SC/ST categories, they ranged from 68.0234447 for OBC, 46.8825338 for SC and 34.6728999 for ST category.

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JEE Main 2022 June and July Session Result - NTA has announced JEE Main result on their website. To download the Scorecard for JEE Main 2022 June and July Session, visit the official website of JEE Main NTA.

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FAQ

1. What is the weightage of the Waves and Oscillations in JEE?

This chapter almost includes 2-3 questions every year and contributes nearly up to 3-4% of the weightage. Therefore, it is important to cover this chapter while preparing for JEE.

2. Is it necessary to study oscillations and waves for the JEE main?

Oscillations and Waves for JEE Main is a critical topic that must be thoroughly studied. Every year, 2-3 questions from this topic appear in the test. These issues are relevant to our daily lives, but we are too distracted to recognise or comprehend them. Therefore, it is necessary to study it.

3. Is it really good to prepare JEE test questions from previous years?

Practicing past year questions from any chapter helps us learn which topics are significant for the exam. It also offers us a sense of the difficulty level of the subjects requested from the chapter. As a result, it is important to practice the previous year's questions for a better comprehension of the chapter's main issues, as well as to try to build your own oscillations and waves notes for the last revision before the test.

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