## NCERT Solutions for Class 11 Physics Chapter 14 OscillationsIn this article we have covered NCERT Solutions for Class 11 Physics Chapter 14 Oscillations. This is one of the most important chapters for any kind of competitive exams as well as for board exams. According to the most recent revision of the CBSE Syllabus 2022-23, the subject specialists developed the solutions. By reading these Solutions from Javatpoint, students can quickly grasp the fundamentals of oscillation. In class 11, students begin to get ready for various competitive exams like JEE Mains and JEE Advanced and these solutions will give them a boost in their preparations. Oscillations is one of the most fundamental chapters of Physic. Referring to the NCERT Solutions offered at Javatpoint will help students comprehend the basic ideas taught in this chapter. An extremely knowledgeable faculty with depth of expertise in the relevant sector develops the answers. According to the weighted marks assigned to each topic in the CBSE test pattern, they make sure to give the students excellent solutions. ## NCERT Solutions for Class 11 Physics Chapter 14
Which of the following examples represents periodic motion? - A swimmer completing one (return) trip from one bank of a river to the other and back.
- A freely suspended bar magnet displaced from its N-S direction and released.
- A hydrogen molecule rotating about its centre of mass.
- An arrow released from a bow.
- The swimmer moves back and forth, but the period isn't fixed. The motion is therefore not periodic.
- The motion of the dangling bar magnet is periodic because it oscillates around its origin every predetermined amount of time.
- The hydrogen molecule rotates around the centre of the mass in an equal-interval periodic motion, returning to its initial location.
- The released arrow does not turn around; it merely goes ahead. This motion is not periodic.
Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? - the rotation of earth about its axis.
- motion of an oscillating mercury column in a U-tube.
- motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
- general vibrations of a polyatomic molecule about its equilibrium position.
- The SHM motion requires a to-and-fro motion, which the earth's rotation does not provide. It is not a SHM motion as a result.
- Since it is a periodic motion, this motion is a SHM motion.
- This oscillatory motion is periodic and periodic about its lower mean position, which is why it is known as SHM motion.
- Although it is not always a SHM, the polyatomic molecule's typical oscillation about its mean position is periodic.
Fig. 14.23 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. 14.23
- Given that there is no repetition and this motion is unidirectional, it cannot be considered periodic.
- It is a periodic motion since the particle's movement repeats every two seconds.
- This motion is not periodic since the particle repeats itself only once, whereas a periodic motion repeats throughout the entire motion.
- Because the particle's motion repeats itself every two seconds, it is a periodic motion.
Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): - sin ωt - cos ωt
- sin
^{3}ωt - 3 cos (π/4 - 2ωt)
- cos ωt + cos 3ωt + cos 5ωt
- exp (-ω
^{2}t^{2}) - 1 + ωt + ω
^{2}t2
a.) The given function is sinωt-cos ωt A Simple Harmonic Motion (SHM) is generally represented by the following equation: b.) The given function is sin Let the given function be f(x) then we can write it as, If we compare equation (iv) with the general equation of SHM (i), then we find that each individual function represents a SHM. However, the superposition of the two waves is not a SHM. Though equation (iv) does not represent a proper equation of SHM, yet the generated wave is periodic in nature. d.) We have given function as cos ωt + cos 3ωt + cos 5ωt Let the given function be considered as f(x). Therefore, f(x) = cos ωt + cos 3ωt + cos 5ωt = cos (ωt + 0 × ∅) + cos (ωt + 3 × ∅) + cos (ωt + 5 × ∅) ...(vii) After comparing equations (vii) and (v), we get to the conclusion that while each individual function reflects a SHM, the superposition of both waves does not. Though not in SHM, the generated wave is periodic in nature. Thus the given function as cos ωt+cos 3ωt + cos 5ωt does not represent a Simple Harmonic Motion. e.) The given function is exp (-ω Let the given function be considered as f(x). Therefore, f(x) = exp (-ω The given function is an exponential function that is being provided. There is no repeated motion around a fixed point in the exponential functions. Since they are not periodic in nature, they do not belong to SHM. However, not all SHM are of a periodic character. Thus the given function as exp (-ω f.) We are given with the function 1 + ωt + ω Let the given function be considered as f(x). Therefore, f(x) = 1 + ωt + ω The given function is quadratic in nature. Unlike motion about a stationary point, the quadratic equation does not recur. They are also not in SHM because they are not periodic in nature. Thus the given function as 1 + ωt + ω
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is: - at the end A
- at the end B
- at the mid-point of AB going towards A
- at 2 cm away from B going towards A
- at 3 cm away from A going towards B
- at 4 cm away from B going towards A
Between points A and B, there is a particle in the SHM. A and B are separated by 10 cm. A direction is positive from A to B. a.) A and B are two end points in the picture below, and O represents their midpoint. At the extreme points A and B, the particle is at rest if it is in SHM between those two positions. Extreme points have zero velocity. At point A, the particle has a tendency to go in the direction of O to keep moving. As a result, O is the direction of acceleration. Positive movement is being made along AB. Positive acceleration is being experienced along AO. As the acceleration is moving in the right direction, force is likewise positive. The velocity at extreme point A is therefore zero. Positive forces and acceleration are present. b.) The graphic below depicts the midpoints of two end points, A and B, as O. At the extreme points A and B, the particle is at rest if it is in SHM between those two positions. Extreme points have zero velocity. The particle starts to move towards O at point B in order to keep moving. As a result, O is the direction of acceleration. Positive movement is being made along AB. Negative acceleration travelling along BO. In line with the direction of the acceleration, force is also negative. The velocity at extreme point B is therefore zero. Negative acceleration and force are present. c.) The graphic below depicts the midpoints of two end points, A and B, as O. There is a particle in SHM. Point O represents the particle's mean location. The greatest velocity is reached at this point. The particle is travelling in a direction opposite to that of B, with a negative velocity. At its mean position, there is no net force. At its mean location, the acceleration is therefore likewise zero. So, point O has the highest velocity. There is no acceleration or force. d.) A and B are two end points with O representing their midpoint in the illustration below From point B, the particle is travelling towards O. Unlike the usual positive direction from A to B, the motion is in the opposite direction. The particle's velocity, acceleration, and force are all negative as a result. e.) A and B are two end points with O representing their midpoint in the illustration below. From point A, the particle is travelling towards O. From point A to point B, motion is in a positive direction. The particle's velocity, acceleration, and force are all positive as a result. f.) A and B are two end points with O representing their midpoint in the illustration below. From point B, the particle is travelling towards O. Unlike the usual positive direction from A to B, the motion is in the opposite direction. The particle's velocity, acceleration, and force are all negative as a result.
Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? - a = 0.7x
- a = -200x
^{2} - a = -10x
- a = 100x
^{3}
We know the condition for SHM which states that, acceleration is inversely related to a particle's displacement. So, if we take a as the acceleration and x as the displacement of a body executing SHM then, we can write the equation, a = -kx (k is any constant) . . . (i) a.) a = 0.7x If we compare the above equation with equation (i) the we notice that they are in different form. Thus we conclude that it is not a SHM. b.) a = -200 x If we compare the above equation with equation (i) the we notice that they are in different form. Thus we conclude that it is not a SHM. c.) a = -10 x If we compare the above equation with equation (i) the we notice that equation are of same type. Thus we conclude that it is a SHM. d.) a = 100 x If we compare the above equation with equation (i) the we notice that they are in different form. Thus we conclude that it is not a SHM.
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ ) If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s
A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?
The scale can read the maximu mass of,M = 50 Kg Maximum displacement of the spring=Length of the scale,l = 20 cm = 0.2 m Time period,T = 0.6 s
A spring having with a spring constant 1200 N m Fig. 14.24 Determine: - the frequency of oscillations
- maximum acceleration of the mass, and
- the maximum speed of the mass.
In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is - at the mean position,
- at the maximum stretched position, and
- 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?
Distance travelled by the mass sideways, a = 2.0 cm Angular frequency of oscillation: (a) As time is noted from the mean position, Using relation,x = a sin ωt we get,x = 2 sin 20 t (b) At maximum stretched position, the body is at the extreme right position, with an initial phase of π / 2 rad. Then, (c) At maximum compressed position, the body is at left position, with an initial phase of 3π / 2 rad. The functions neither differ in amplitude nor in frequency. They differ in initial phase.
Figures 14.25 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Figures 14.25 Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.
a.) Time period, t = 2 s Amplitude, A = 3 cm At time, t = 0, the radius vector OP makes an angle π / 2 with the positive x -axis, it means that, Phase angle ϕ = + π / 2 Hence, the equation of simple harmonic motion for the x - projection of OP, at the time t, is given by the displacement equation: Thus, x = - 3 sin (πt) cm b.) Time period, t = 4 s Amplitude, a = 2 m At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction Thus, Phase angle ϕ = +π Hence, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given as: x = - 2 cos {(π / 2) t}m
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
Figure (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. (b) is stretched by the same force F. (a) What is the maximum extension of the spring in the two cases? (b) If the mass in Fig. (a) and the two masses in Fig. (b) is released, what is the period of oscillation in each case?
(a) The maximum extension of the spring in fig (a) is x. The reaction force acting on the other mass is created by the force on each mass. The two masses act as though they are fixed in relation to one another. (b) In figure (a) the restoring force on the mass is F = -kx, here x is the extension of the spring. For the mass (m) of the block, force is written as
The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
The acceleration due to gravity on the surface of moon is 1.7 m s
Acceleration due to gravity on the surface of moon,g' = 1.7 m s Acceleration due to gravity on the surface of earth,g = 9.8 m s Time period of a simple pendulum on earth,T = 3.5 s Therefore, the simple pendulum's time period on the moon's surface is 8.4 seconds.
Answer the following questions: a.) Time period of a particle in SHM depends on the force constant k and mass m of the particle: A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum? b.) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than Think of a qualitative argument to appreciate this result. c.) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give the correct time during the free fall? d.) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?
a.) The spring constant k for a straightforward pendulum is inversely proportional to mass. The denominator and the m, which is the numerator, will cancel each other out. Consequently, the mass of the bob has no effect on the simple pendulum's time period. b.) The term describes the restoring force acting on a basic pendulum's bob. F = -mg sin θ Where, F is the restoring force m is the mass of the bob g is the acceleration due to gravity θ is the angle of displacement When θ is small,sin θ ≈ θ. Then the expression for the time period of a simple pendulum is given by When θ is large, sin θ < θ. Therefore, the above equation will not be valid. There will be an increase in the time period T. (c) A wristwatch operates on spring action rather than gravity's acceleration. As a result, the watch will display the accurate time. (d) The cabin's acceleration owing to gravity will be zero during free fall. As a result, the basic pendulum's frequency of oscillation will also be zero.
A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
The circular motion of the car plus the acceleration brought on by gravity will cause the bob of the simple pendulum to experience centripetal acceleration. Acceleration due to gravity = g Where, v is the uniform speed of the car R is the radius of the track Effective acceleration (g') is given as
A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period where ρ is the density of cork. (Ignore damping due to viscosity of the liquid)
Base area of the cork = A Height of the cork = h Density of the liquid = ρ Density of the cork = ρ In equilibrium: Weight of the cork = Weight of the liquid displaced by the floating cork Let the cork be displaced by a small distance x Therefore,some excess water will be displaced of a c ertain volume As a result, an additional upward thrust acts and gives the cork the restoring force. Upthrust = restoring force,F = weight of the extra water displaced F = -(volume × density × g) Volume = area×distance through which the cork is displaced Volume = Ax Therefore,F = -Ax × ρ According to the force law: Where, m = mass of the cork = volume of the cork × density = base area of the cork × height of the cork × density ofthe cork = Ahρ Therefore, the expression for the time period becomes:
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.
Area of cross - section of the U - tube = A Density of the mercury column = ρ Acceleration due to gravity = g Let the restoring force be F ∴F = Weight of the mercury column of a certain height F = -(volume × density × g) F = - (A × 2h × ρ × g) = -2Aρgh = -k × displacement in one of the arms (h) Where, 2h is the height of the mercury column in the two arms k is a constant, given by, Where, m is the mass of the mercury column Let l be the length of the total mercury in the U - tube Mass of mercury,m = volume of mercury x Density of mercury = Alρ Hence, Therefore, the mercury column executes simple harmonic motion with time period
An air chamber of volume V has a neck area of cross-section into which a ball of mass m just fits and can move up and down without any friction. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Figure].
Volume of the air chamber = V Cross-sectional area of the neck = A Mass of the ball = m At position C, the ball is inserted into the neck. In the chamber, the air pressure beneath the ball is equivalent to the atmospheric pressure. By slightly increasing the pressure (by a small amount p), the ball is forced downward and advances to position D. The distance CD =y The air chamber's volume shrinks as its pressure rises. The volume of the air inside the chamber will decrease, resulting in a rise in pressure. Air volume inside the chamber decreases, From equation (i) we can see that F ∝ y and the negative sign indicates that the force is directed towards the equilibrium position. The ball will move in a simple harmonic motion with C as the mean position in the neck of the chamber if the increased pressure is eliminated. In S.H.M.,the restoring force is given by,F = -ky . . . (ii) On comparing equation (i) and (ii)
You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.
The mass of the automobile is = 3000 Kg The displacement is = 15 cm Each wheel supports a mass = 750 Kg After each complete oscillation,the amplitude decreases by = 50 % a.) Let k be the spring constant of one spring. Thus,total spring constant of four springs would be = 4k in parallel
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.
Let the mass of the particle executing simple harmonic motion be m. Thus, the displacement of the particle at any instant t will be given by the SHM equation.
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = -α θ, where J is the restoring couple and θ the angle of twist).
Mass of the circular disc = 10 Kg Period of torsional oscillation = 1.5 s Radius of the disc=15 cm = 0.15 m Restoring couple,J = -αθ
A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm.
Given amplitude of the body,A = 5 cm = 00.5 m Time period of the body is,T = 0.2 s a.) For displacement,x = 5 cm = 0.05 m Acceleration is given by: b.) For displacement,x=3 cm=0.03 m Acceleration is given by: c.) For displacement,x = 0 Acceleration is given by:
A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x
According to problem, for an oscillating mass the displacement equation is given by, x = A cos (ωt + θ) Where, A is the amplitude x is the displacement θ is the phase constant |