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Have you seen a swing going back and forth or a ball bouncing? These are oscillations. In this chapter, you will study how objects move in repeating patterns, such as a pendulum or the vibration of a guitar string. Oscillations are everywhere, from the ticking to sound waves' vibrations.
Oscillations is a crucial topic in the Class 11 NCERT syllabus, covering fundamental concepts. The Oscillations chapter in Class 11 Physics is crucial as it forms the foundation for wave motion, SHM (Simple Harmonic Motion), and resonance. It is widely applied in pendulums, springs, sound waves, AC circuits, and quantum mechanics. If you struggle to grasp the concept of Oscillations or solve NCERT textbook questions, Careers360 provides detailed NCERT solutions to assist you in your studies.
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Through these problems, the students gain a robust concept of wave motion, mechanical vibrations, and oscillatory systems. All of these topics are crucial for CBSE exams, JEE Main, and NEET since oscillation is an important chapter.
Searching for a simple and free solution to crack your Class 11 Physics exams? You can download the Chapter 13 NCERT Solutions PDF. It's full of step-by-step solutions to help you master waves and vibrations and make complex material easy and simple to understand. Ideal for last-minute revisions and to shine in your exam.
Q. 13.1 Which of the following examples represent periodic motion?
(a) A swimmer completing one (return) trip from one bank of a river to the other and back.
(b) A freely suspended bar magnet displaced from its N-S direction and released.
(c) A hydrogen molecule rotating about its centre of mass.
(d) An arrow released from a bow
Answer:
(a) The motion is not periodic, though it is to and fro.
(b) The motion is periodic.
(c) The motion is periodic.
(d) The motion is not periodic.
Q. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of the earth about its axis.
(b) motion of an oscillating mercury column in a U-tube.
(c) motion of a ball bearing inside a smooth curved bowl when released from a point slightly above the lowermost point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.
Answer:
(a) Periodic but not S.H.M.
(b) S.H.M.
(c) S.H.M.
(d) Periodic but not S.H.M.M [A polyatomic molecule has a number of natural frequencies, so its vibration is a superposition of SHM’s of a number of different frequencies. This is periodic but not SHM]
Q. 13.3 Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?
Fig 13.18
Answer:
The x-t plots for linear motion of a particle in Fig. 13.18 (b) and (d) represent periodic motion with both having a period of motion of two seconds.
Q. 13.4 (a) Which of the following functions of time represent
(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion (
(a)
Answer:
Since the above function is of form
(b)
Answer:
The two functions individually represent SHM but their superposition does not give rise to SHM the motion will definitely be periodic with a period of
(c)
Answer:
The function represents SHM with a period of
(d)
Answer:
Here, each individual function is SHM. But superposition is not SHM. The function represents periodic motion but not SHM.
(e)
Answer:
The given function is exponential and, therefore, does not represent periodic motion.
(f)
Answer:
The given function does not represent periodic motion.
(a) at the end A,
Answer:
Velocity is zero. Force and acceleration are in the positive direction.
(b) at the end
Answer:
Velocity is zero. Acceleration and force are negative.
(c) at the mid-point of AB going towards
Answer:
Velocity is negative, that is, towards A, and its magnitude is maximum. Acceleration and force are zero.
(d) at
Answer:
The velocity is negative. Acceleration and force are also negative.
(e) at
Answer:
Velocity is positive. Acceleration and force are also positive.
(f) at
Answer:
Velocity, acceleration and force all are negative
(a)
(b)
(c)
(d)
Answer:
Only the relation given in (c) represents simple harmonic motion as the acceleration is proportional in magnitude to the displacement from the midpoint, and its direction is opposite to that of the displacement from the mean position.
If the initial
Answer:
at t = 0
at t = 0
Squaring and adding equation (i) and (ii), we get
Dividing equation (ii) by (i) we get
at t = 0
at t = 0
Squaring and adding equation (iii) and (iv), we get
Dividing equation (iii) by (iv), we get
Answer:
Spring constant of the spring is given by
The time period of a spring attached to a body of mass m is given by
Fig 13.19
Determine
(i) the frequency of oscillations,
Answer:
The frequency of oscillation of an object of mass m attached to a spring of spring constant k is given by
Fig 13.19
Determine
(ii) maximum acceleration of the mass, and
Answer:
A body executing S.H.M experiences maximum acceleration at the extreme points
Fig 13.19
Determine
(iii) the maximum speed of the mass.
Answer:
Maximum speed occurs at the mean position and is given by
(a) at the mean position,
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer:
Amplitude is A = 0.02 m
Time period is
(a) At t = 0 the mass is at mean position i.e. at t = 0, x = 0
Here x is in metres and t is in seconds.
(b) at the maximum stretched position,
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer:
Amplitude is A = 0.02 m
Time period is
(b) At t = 0 the mass is at the maximum stretched position.
x(0) = A
Here x is in metres and t is in seconds.
(c) at the maximum compressed position.
In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?
Answer:
Amplitude is A = 0.02 m
Time period is
(c) At t = 0 the mass is at the maximum compressed position.
x(0) = -A
Here x is in metres and t is in seconds.
The above functions differ only in the initial phase and not in amplitude or frequency.
Fig 13.20
Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:
(a) Let the required function be
Amplitude = 3 cm = 0.03 m
T = 2 s
Since initial position x(t) = 0,
As the sense of revolution is clockwise
Here x is in metres and t is in seconds.
(b)Let the required function be
Amplitude = 2 m
T = 4 s
Since initial position x(t) = -A,
As the sense of revolution is anti-clockwise
Here x is in metres and t is in seconds.
(a)
Answer:
The initial position of the particle is x(0)
The radius of the circle i.e. the amplitude, is 2 cm
The angular speed of the rotating particle is
The initial phase is
The reference circle for the given simple Harmonic motion is
(b)
Answer:
The initial position of the particle is x(0)
The radius of the circle i.e. the amplitude is 1 cm
The angular speed of the rotating particle is
Initial phase is
The reference circle for the given simple Harmonic motion is
(c)
Answer:
At t= 0
The reference circle is as follows
(d)
Answer:
The initial position of the particle is x(0)
The radius of the circle i.e. the amplitude is 2 cm
The angular speed of the rotating particle is
Initial phase is
The reference circle for the given simple Harmonic motion is
(b) is stretched by the same force F.
Fig 13.21
(a) What is the maximum extension of the spring in the two cases?
Answer:
(a) Let us assume the maximum extension produced in the spring is x.
At maximum extension
(b) Let us assume the maximum extension produced in the spring is x. That is x/2 due to force towards left and x/2 due to force towards right
Answer:
Amplitude of SHM = 0.5 m
angular frequency is
If the equation of SHM is given by
The velocity would be given by
The maximum speed is therefore
Answer:
The time period of a simple pendulum of length l executing S.H.M is given by
g e = 9.8 m s -2
g m = 1.7 m s -2
The time period of the pendulum on the surface of Earth is T e = 3.5 s
The time period of the pendulum on the surface of the moon is T m
Answer:
Acceleration due to gravity = g (in downward direction)
Centripetal acceleration due to the circular movement of the car = a c
Effective acceleration is
The time period is T'
Show that the cork oscillates up and down simply harmonically with a period
Answer:
Let the cork be displaced by a small distance x in downwards direction from its equilibrium position where it is floating.
The extra volume of fluid displaced by the cork is Ax
Taking the downwards direction as positive we have
Comparing with a=-kx we have
Answer:
Let the height of each mercury column be h.
The total length of mercury in both columns = 2h.
Let the cross-sectional area of the mercury column be A.
Let the density of mercury be
When either of the mercury columns dips by a distance x, the total difference between the two columns becomes 2x.
The weight of this difference is
This weight drives the rest of the entire column to the original mean position.
Let the acceleration of the column be a. Since the force is restoring
The time period of the oscillation would be
Fig 1
Answer:
Let the initial volume and pressure of the chamber be V and P.
Let the ball be pressed by a distance x.
This will change the volume by an amount ax.
Let the change in pressure be
Let the Bulk's modulus of air be K.
This pressure variation would try to restore the position of the ball.
Since force is restoring in nature, displacement and acceleration due to the force would be in different directions.
The above is the equation of a body executing S.H.M.
The time period of the oscillation would be
(a) the spring constant
Answer:
Mass of automobile (m) = 3000 kg
There are a total of four springs.
Compression in each spring, x = 15 cm = 0.15 m
Let the spring constant of each spring be k
(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports
Answer:
The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.
For damping factor b we have
x=x 0 /2
t=0.77s
m=750 kg
Answer:
Let the equation of oscillation be given by
Velocity would be given as
Kinetic energy at an instant is given by
Time Period is given by
The Average Kinetic Energy would be given as follows
The potential energy at an instant T is given by
The Average Potential Energy would be given by
We can see K av = U av
Answer:
Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is
The period of Torsional oscillations would be
(a)
Answer:
A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 5 cm
(b)
Answer:
A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 3 cm
(c)
Answer:
A = 5 cm = 0.05 m
T = 0.2 s
At displacement x acceleration is
At displacement x velocity is
(a)At displacement 0 cm
Answer:
At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring.
Let the amplitude be A
The angular frequency of a spring-mass system is always equal to
Therefore
NCERT answers for oscillations allow students to have a thorough understanding of the various factors that affect an object's motion. Period, frequency, simple harmonic motion, and uniform circular motion are a few of these factors. Chapter 13 of Physics in Class 11 NCERT Solutions offers a thorough explanation of every important concept involved, such as force laws, velocity, forced oscillations, systems that undergo harmonic motion (such as spring- mass systems), etc.
Have you ever wondered why gases appear to expand or how they produce pressure? The Kinetic Theory of Gases explains it by imagining that gases consist of small particles moving about rapidly. Scientists such as Boyle, Newton, Maxwell, and Boltzmann contributed significantly to this theory, which explains gas behavior, their response to changes in temperature and pressure, and even how they flow and mix together.
NCERT solutions for important chapter kinetic theory class 11 chapter 12 physics are created by subject matter experts to offer accurate and clear answers to each and every NCERT exercise problem. Students can build a strong concept by practicing these Kinetic Theory Class 11 Physics solutions. Step-by-step solutions make learning convenient and are provided below.
Also Read
Free download of Kinetic Theory of Gases Class 11 Solutions PDF for CBSE exams and strengthen your understanding with step-by-step explanations and solved exercises.
Answer:
The diameter of an oxygen molecule, d = 3 Å.
The actual volume of a mole of oxygen molecules V actual is
The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres
Answer:
As per the ideal gas equation
For one mole of a gas at STP we have
Q12.3 Figure 13.8 shows plot of
(a) What does the dotted plot signify?
(b) Which is true:
(c) What is the value of
(d) If we obtained similar plots for
value of
of hydrogen yields the same value of
region of the plot) ? (Molecular mass of
Answer:
(a) The dotted plot corresponds to the ideal gas behaviour.
(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T 1 is closer to the horizontal line that the one corresponding to T 2 we conclude T 1 is greater than T 2.
(c) As per the ideal gas equation
The molar mass of oxygen = 32 g
R = 8.314
(d) If we obtained similar plots for
value of
Molar Mass of Hydrogen M = 2 g
mass of hydrogen
Answer:
Initial volume, V 1 = Volume of Cylinder = 30 l
Initial Pressure P 1 = 15 atm
Initial Temperature T 1 = 27 o C = 300 K
The initial number of moles n 1 inside the cylinder is
Final volume, V 2 = Volume of Cylinder = 30 l
Final Pressure P 2 = 11 atm
Final Temperature T 2 = 17 o C = 290 K
The final number of moles n 2 inside the cylinder is
Moles of oxygen taken out of the cylinder = n 2 -n 1 = 18.28 - 13.86 = 4.42
Mass of oxygen taken out of the cylinder m is
Answer:
Initial Volume of the bubble, V 1 = 1.0 cm 3
Initial temperature, T 1 = 12 o C = 273 + 12 = 285 K
The density of water is
Initial Pressure is P 1
Depth of the bottom of the lake = 40 m
Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K
Final Pressure = Atmospheric Pressure
Let the final volume be V 2
As the number of moles inside the bubble remains constant, we have
Answer:
The volume of the room, V = 25.0 m 3
Temperature of the room, T = 27 o C = 300 K
The pressure inside the room, P = 1 atm
Let the number of moles of air molecules inside the room be n
Avogadro's Number,
The number of molecules inside the room is N
Answer:
The average energy of a Helium atom is given as
(i)
(ii)
(iii)
Answer:
As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.
Root mean square velocity is given as
As we can see, vrms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon as its molar mass is the least.
Answer:
As we know root mean square velocity is given as
Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at
Answer:
Pressure, P = 2atm
Temperature, T = 17 o C
The radius of the Nitrogen molecule, r=1 Å.
The molecular mass of N 2 = 28 u
The molar mass of N 2 = 28 g
From the ideal gas equation
The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as
The mean free path
The root mean square velocity v rms is given as
The time between collisions T is given as
Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter
The ratio of the average time between collisions to the collision time is
Thus, we can see that the time between collisions is much larger than the collision time.
Answer:
Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P 1 = 1 atm = 76 cm of Mercury
Let the crossectional area of the tube be x cm 2
The initial volume of the air column, V 1 = 15x cm 3
Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.
The pressure of the air column after y cm of Mercury has flown out of the column P 2 = 76 - (76 - y) cm of Mercury = y cm of mercury
Final volume of air column V 2 = (24 + y)x cm 3
Since the temperature of the air column does not change
Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm
Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.
Length of the air column = y + 24 = 47.8 cm.
Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm
Answer:
As per Graham's Law of diffusion, if two gases of Molar Mass M 1 and M 2 diffuse with rates R 1 and R 2 respectively, their diffusion rates are related by the following equation
In the given question
R 1 = 28.7 cm 3 s -1
R 2 = 7.2 cm 3 s -1
M 1 = 2 g
The above Molar Mass is close to 32, therefore, the gas is Oxygen.
Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres
where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column:
where
Answer:
Let the suspended particles be spherical and have a radius r
The gravitational force acting on the suspended particles would be
The buoyant force acting on them would be
The net force acting on the particles become
Replacing mg in equation (i) with the above equation, we get
The above is the equation to be derived
Substance |
Atomic Mass (u) |
Density (10 3 Kg m 3 ) |
Carbon (diamond) |
12.01 |
2.22 |
Gold |
197 |
19.32 |
Nitrogen (liquid) |
14.01 |
1 |
Lithium |
6.94 |
0.53 |
Fluorine |
19 |
1.14 |
Answer:
Let one mole of a substance of atomic radius r and density
Let us assume the atoms to be spherical
Avogadro's number is
For Carbon
For gold
For Nitrogen
For Lithium
For Fluorine
Class 11 Physics Chapter 12 NCERT solutions help students understand how tiny particles in matter behave. This chapter explains gases and their thermal properties, linking to thermodynamics. Studying it helps in exams and builds a strong foundation in physics.
12.1 Introduction
The kinetic theory explains the behaviour of gases by assuming they consist of a large number of tiny particles (atoms or molecules) in constant random motion. This theory helps derive gas laws and explain properties like pressure, temperature, and specific heat.
12.2 Molecular Nature of Matter
12.3 Behavior of Gases
where
12.4 Kinetic Theory of an Ideal Gas
where
where
12.5 Law of Equipartition of Energy
where
12.6 Specific Heat Capacity
12.7 Mean Free Path
where
Section | Topic Name |
12.1 | Introduction |
12.2 | Molecular nature of matter |
12.3 | Behaviour of gases |
12.4 | Kinetic theory of an ideal gas |
12.5 | Law of equipartition of energy |
12.6 | Specific heat capacity |
12.7 | Mean free path |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 |
For JEE Mains, oscillation and wave questions represent 6.67 percent of the total questions. The majority of the earlier oscillation-related JEE mains problems were from the subjects of SHM and basic pendulum. There will likely be two oscillation-related questions in the NEET test. To perform well in Class 11 and competitive exams, use the CBSE NCERT answers for Class 11 Physics Chapter 13 Oscillations.
1. Understand Basics First- Understand what oscillations and SHM are before diving into formulas. Focus on SHM Concepts. Learn words such as amplitude, frequency, period, and phase correctly.
2. Use Graphs & Diagrams- Visualize SHM using displacement, velocity, and acceleration vs time graphs.
3. Memorize Formulas with Meaning- Don't just memorize—understand what each formula means and when to apply it.
4. Practice NCERT Numericals- Complete all examples and exercises, even for solid conceptual use.
5. Revise Regularly- Create a formula sheet and regularly revise it.
NCERT solutions for class 11 Subject-wise
From the NCERT chapter oscillations, two questions can be expected for NEET exam. For more questions solve NEET previous year papers.
Yes, oscillation is important for JEE Main. One or two question can be expected from oscillations for JEE Main. It is one of the important chapter for scholarship exams like KVPY and NSEP. To solve more problems on Oscillations refer to NCERT book, NCERT exemplar and JEE main previous year papers.
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