NCERT Solutions for Class 11 Physics Chapter 14

NCERT Solutions for Class 11 Physics Chapter 14: Oscillations

Class 11 Physics Chapter 14 is one of the most popular topics in the Physics term – II examination is oscillations. Subject specialists at Extramarks have prepared the solutions in accordance with the most recent update on CBSE Syllabus 2022-23 term-by-term. Students may easily grasp the ideas of Oscillation by consulting Extramarks NCERT Solutions.

Students in Class 11, begin to study for competitive exams like JEE, NEET, CLAT, AIIMS, and others. Physics is a  part of these crucial exams , and it demands extensive notes and preparatory help.The ideal strategy to reach the goal of performing well in various exams is to use NCERT Solutions not just for Physics but for other core subjects as well. 

Oscillatory motion is a fundamental concept in physics. The NCERT Solutions on Extramarks can help students understand the essential concepts discussed in this chapter. The solutions are prepared by a team of highly qualified experts with extensive experience in the sector. They ensure that students receive high-quality solutions based on the weighted marks assigned to each topic in the CBSE term – II exam pattern. Here you can access the NCERT Solutions for Class 11 Physics Chapter 14 for better learning and scoring good grades.

NCERT Solutions for Class 11 Physics Chapter 14: Oscillations

These NCERT Solutions include answers to textbook questions, noteworthy questions from previous year’s exam papers, and sample papers. Worksheets, sample problems, short and long answer questions, MCQs, and tips and tricks are included in the solution to assist the students to  understand and prepare them for any kind of exam. NCERT Solutions are one of the greatest resources for preparing for Class 11 Physics and oscillations is a crucial chapter in the CBSE Class 11 Physics term – II Syllabus. Students must study this chapter thoroughly in order to achieve excellent grades on their second-term exams. The NCERT Solutions for  Oscillations are provided with additional exercises so that students may fully comprehend the concepts covered in this Chapter.

NCERT Physics Class 11 Oscillations

The NCERT Solutions for Class 11 Physics Chapter 14 covers twenty-five questions based on the rules of oscillation and waves, as per the CBSE syllabus.

Class 11 Physics Chapter 14: Chapter Summary & Sub-topics

Oscillatory motion deals with an object’s to and fro motion from its initial position. In the absence of friction, an item can remain in oscillatory motion indefinitely, however, this is not possible due to equilibrium.

In addition, Galileo found four major moons of Jupiter in 1610.  Each moon appeared to move back and forth relative to the planet in a simple harmonic motion to him, with the planet’s disc serving as the motion’s halfway. The handwritten record of his observations is still available.

Galileo used his pulse beats to estimate the periods of a swinging chandelier in a cathedral.  He saw that the chandelier’s movement was regular. The system is similar to  a pendulum. Free oscillations can be seen in a person swinging without being pushed or in a  simple pendulum that has been pushed and released. In both cases,   the swing amplitude will gradually decrease, and the system will eventually come to a standstill. The free oscillations cannot be sustained in practice due to the ever-present dissipative forces. 

The phenomenon of amplitude increasing when the driving power is close to the oscillator’s natural frequency is known as resonance. We encounter resonance-related events in our daily lives.

Your swinging experience is a good example of resonance. You may have realised that the key to swinging to greater heights is to synchronise the rhythm of pushing against the ground with the natural frequency of the swing. The real oscillator and its motion are then described as damped.

A combination of two simple harmonic motions with arbitrary amplitudes and phases is not always periodic. It’s only periodic if one motion’s frequency is an integral multiple of the other’s frequency. A periodic motion, on the other hand, may always be described as the sum of an unlimited number of harmonic motions of varying amplitudes.

NCERT Class 11 Physics Chapter 14 subtopics are listed below.

  1. Introduction
  2. Periodic and oscillatory motions
  3. Simple harmonic motion
  4. Simple harmonic motion and uniform circular motion
  5. Velocity and acceleration in simple harmonic motion
  6. Force law for simple harmonic motion
  7. Energy in simple harmonic motion
  8. Some systems executing SHM
  9. Damped simple harmonic motion
  10. Forced oscillations and resonance

To do well on their term – II examination, students should become familiar with the principles in this chapter. Students will find the principles in this module beneficial while preparing for competitive admission examinations such as JEE and NEET.

While studying for their physics term exams, students must use different strategies. . The following are some effective preparation methods. 

  • During their preparation, students must be fully aware of the most recent CBSE Syllabus 2022-23 by term. It will  help them to know the first and second-term test patterns.
  • During their test preparation, they must stick to a schedule.
  • While studying, students must be thorough with the NCERT books to excel in their exams.
  • Taking notes is one of the most effective strategies to remember things for a longer period of time.
  •  Practice with a variety of question papers and sample papers after going through the NCERT books.

Extramarks provides you with the best study tools, notes, sample papers, important questions, MCQs, NCERT Books, and tips & tricks to help you to take the Class 11 term I and II examinations confidently. The videos and animations created by Extramarks on key topics in every subject make it easier to understand and easily remember those concepts  for a long time.

A Quick Glance  through The First Ten Questions in The Solution

There are twelve other theoretical and equation-based exercises  in Class 11 Physics Chapter 14 NCERT Solutions in addition to these ten questions. Students must use pictures and situations to answer issues and determine equations. Solving these difficult questions will build the physics foundation needed for  higher studies.

Why is NCERT Solutions Class 11 Physics Oscillations PDF the Best Resource for Students?

Young students can prepare for competitive exams by reading the solutions from Oscillations and Waves Class 11 NCERT Solutions. Apart from enhancing a student’s preparation demands, it is advantageous in the following ways:

  • These NCERT Solutions for Class 11 Physics Chapter 14 Oscillations include step-by-step explanations.
  • Reading the answers will clarify the learner’s understanding of important concepts such as oscillation and waves.
  • These answers were curated by subject matter experts and are ideal for CBSE and competitive exams.
  • These solutions are written in a simple and precise manner to ensure good exam results.

A student might seek guidance from online education sources in addition to reading the NCERT Solutions for Class 11 Physics Chapter 14. With so many alternatives, Extramarks NCERT Solutions is the most suitable and the best option for a dedicated and diligent student to achieve a high score. It’s just a matter of making the right decision, at the right time  and to find the right source to learn and practise from. Rest is easy. Extramarks Solutions is the master key to all your subjects- it’s simple, easy and accessible.

NCERT Solutions for Class 11 Physics Chapter 14 

NCERT Solutions for Class 11 Physics Chapter 14 have answers to all the questions given at the end of Chapter 14 in NCERT book. Students can refer to these solutions to solve questions as well as cross-check their answers.

Q.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 center of mass.
(d) An arrow released from a bow.

Ans.

(a) The motion of the swimmer is not periodic as the swimmer’s motion is only back and forth. Its time period is also not definite as the time taken by the swimmer during his back and forth journey may not be the same.

(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released represents periodic motion as a freely suspended magnet oscillates about its position with a specific time period.

(c) When the hydrogen molecule rotating about its centre of mass represents a periodic motion as it reaches the same position after a fixed time again and again.

(d) An arrow released from a bow travels only in the forward direction so it does not represent a periodic motion.

Q.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(a) the rotation of 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 lower most point.
(d) general vibrations of a polyatomic molecule about its equilibrium position.

Ans.

(a) During the rotation of earth about its axis, the earth reaches the same position after fixed duration of time. Therefore, it represents a periodic motion. But, it is not a simple harmonic motion, because earth does not have a to and fro motion about its axis.

(b) The motion of an oscillating mercury column in a U-tube represents a simple harmonic motion. This is due to the fact that the mercury moves to and fro on the same path, about the fixed position, with a definite time period.

(c) The ball comes back to its initial position in the same period of time, again and again moving in a to and fro motion. Therefore, its motion is a periodic and simple harmonic

(d) The polyatomic molecule has many natural frequencies of vibrations. Its general motion is the superposition of individual simple harmonic motions of different molecules. Thus, it represents the periodic motion but not the simple harmonic motion.

Q.3 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)?

Ans.

(a) It does not represent a periodic motion, because there is no repetition of motion in this case.

(b) In the given X-t plot, the motion of the particle repeats itself after every 2 s. Therefore, it is surely a periodic motion, whose period is 2 s.

(c) The given X-t plot does not represent a periodic motion. This is due to the fact that the particle repeats the motion in one position only, whereas, in case of a periodic motion, the entire motion of the particle must be repeated periodically.

(d) In the given X-t plot, the motion of the particle repeats itself after 2 s. Therefore, it represents a periodic motion having a time period of 2 s.

Q.4

Which of the following functions of timerepresent ( a ) simple harmonic, ( b ) periodicbut not simple harmonic, and ( c ) non-periodicmotion? Give periodfor each case of periodicmotion (ω is any positive constant): ( a ) sinωt cosωt ( b ) sin 3 ωt ( c ) 3 cos ( π 4 2ωt ) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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ndaaaGccqaHjpWDcaqG0baabaWaaeWaaeaacaqGJbaacaGLOaGaayzkaaGaaeiiaiaabodacaqGGaGaae4yaiaab+gacaqGZbGaaeiiamaabmaabaWaaSaaaeaacqaHapaCaeaacaqG0aaaaiaabccacaGGtaIaaeiiaiaabkdacqaHjpWDcaqG0baacaGLOaGaayzkaaaaaaa@1CD1@ ( d ) cos ωt+ cos 3ωt+ cos 5ωt ( e ) exp ( ω 2 t 2 ) ( f ) 1 +ωt+ ω 2 t 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@68B4@

Ans.

( a )Here,thegivenfunction is: sinωtcosωt= 2 [ 1 2 sinωt 1 2 cosωt ] = 2 sin( ωt π 4 ) As thegiven functioncan be written in the form: asin( ωt+ϕ ),therefore,it represents SHM Period ofthegiven function= 2π ω ( b ) Here,the given function is: sin 3 ωt= 1 4 [ 3sinωtsin3ωt ] Here,the terms sinωtand sin3ωtrepresent simpleharmonic motion ( SHM )individually.But, the superposition of two SHM is periodic and not simpleharmonic. ( c )Here,the given function is: 3cos[ π 4 2ωt ]=3cos[ 2ωt π 4 ] Thegiven function represents SHM. Period ofthegiven function= 2π 2ω = π ω ( d )The given function=cosωt+cos3ωt+cos5ωt Inthiscase,each individual cosine functionrepresentsSHM. Here, the superposition of three SHM isperiodic,but not simple harmonic. ( e ) The given function=exp( ω 2 t 2 ) Itis an exponential function. Astheexponentialfunctionsdonot repeat themselves,therefore, it does notrepresentaperiodicmotion. ( f ) The given function=1+ωt+ ω 2 t 2 Itrepresentsthenonperiodicmotion. 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Q.5 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
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A

Ans.

The situation of the problem is shown in the given figure. The particle is in linear SHM. Here, points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path AB.

(a) In this case, the particle is at the extreme point A, at rest momentarily. Therefore, its velocity is zero at this point.

The acceleration of the particle is positive as its direction is along AO.

As the particle is directed rightward, therefore, force is also positive in this case.

(b) In this case, the particle is at the extreme point B, at rest momentarily. Therefore, the velocity of the particle is equal to zero at this point.

As it is directed along B, therefore, its acceleration is negative.

As the particle is directed leftward, therefore, force is also negative in this case.

(c)

In this case, the particle is in simple harmonic motion. Here, O represents the mean position of the particle. At the mean position O, the velocity of the particle is the maximum. As the particle is directed leftward, therefore, the velocity is negative. At the mean position O, the acceleration and force of a particle executing SHM is zero.

(d)

The particle is traveling toward point O from the end B. In this case, the direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, in this case, the velocity and acceleration of the particle are negative. The force on it is also negative.

(e)

In this case, the particle is traveling towards point O from the end A. Here, the direction of motion is from A to B, which is the positive direction conventionally. Therefore, the velocity, acceleration, and force are all positive in this case.

(f)

This case is similar to the case given in (d). Therefore, velocity, acceleration and force are all negative in this case.

Q.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = –200x2
(c) a = –10x
(d) a = 100x3

Ans.

In the SHM, acceleration is related to displacement as:

a = –kx

Out of the given relations, only relation (c) is written in the above form. Therefore, relation (c) represents the SHM.

Q.7 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 cms-1, what are its amplitude and initial phase angle? The angular frequency of the particle is πs–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Ans.

Here,initially, att=0: Displacement,x=1 cm Initial velocity,v=ω cm sec -1 Angular frequency, ω=π rads 1 Given that: x( t )=Acos( ωt+ϕ ) 1=Acos( ω×0+ϕ )=Acosϕ Acosϕ=1(i) Asvelocity,v= dx dt ω=Aωsin( ωt+ϕ ) 1=Asin( ω×0+ϕ )=Asinϕ Asinϕ=1(ii) Squaring and adding equations ( i ) and ( ii ), we have: tanϕ=1 ϕ= 3π 4 , 7π 4 ,….. SHMisgivenbytherelation: x=Bsin( ωt+α ) Puttingthegivenvaluesintheaboverelation,weobtain: 1=Bsin[ ω×0+α ] Bsinα=1(iv) Squaringandaddingequation(iii)and(iv),wehave: B 2 [ sin 2 α+ cos 2 α ]=1+1 B 2 =2 B= 2 cm Dividing equation ( iii ) by equation ( iv ), we obtain: Bsinα Bcosα = 1 1 tanα=1=tan π 4 α= π 4 , 5π 4 ,…….. 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Q.8 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?

Ans.

Here,maximum mass that the scale can read,M = 50 kg Maximum extension of spring = Length of scale, l = 20 cm = 0.2 m Time periodofoscillation,T = 0.6 s Maximum force applied on the spring, F = Mg Here,g = acceleration due to gravity = 9 .8 ms -2 F = 50 kg× 9 .8 ms -2 = 490 N Spring constant,k = F l = 490 N 0.2 m = 2450 Nm -1 Letmass suspended from the balance be m. Time period,T = 2π m k m = T 2 ×k = 0.6 s 2×3.14 2 ×2450 Nm -1 = 22.36kg Weight of body,W = mg = 22.36×9.8 = 219.167 N The weight of the body is approximately 219 N.

Q.9 A spring having with a spring constant 1200 Nm–1 is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Ans.

Here,spring constant, k = 1200 Nm –1 Massattached,m = 3 kg Displacementofthemass,A = 2.0 cm = 0.02 m i Frequency of oscillation, ν = 1 T = 1 k m Here,T= Time period ν = 1 2×3.14 1200 Nm -1 3 kg = 3.18 ms -1 The frequency of oscillations is 3 .18 ms -1 . ii Maximum acceleration, a = ω 2 A Here, ω = Angular frequency = k m A = Maximum displacement a = k m A = 1200 Nm -1 ×0.02 m 3 kg = 8 ms -2 The maximum acceleration of the mass is 8 .0 ms -2 . iii Maximum velocity, v max = Aω = A k m v max = 0.02 m× 1200 Nm -1 3 kg = 0.4 ms -1 The maximum velocity of the mass is 0 .4 ms -1 .

Q.10 In Exercise 14.9, let us take the position of mass when the spring is unstretched 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
(a) at the mean position,
(b) at the maximum stretched position, and
(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?

Ans.

Here,distance travelled by the mass,a=2.0 cm Springconstant,k= 1200 Nm –1 Massattached,m=3 kg Angular frequency of oscillationisgivenas: ω= k m = 1200 3 = 400 = 20 rad s –1 ( a )When the mass is at the mean position,Initial phase=0 Displacementisgivenas: x=Asinωt =2sin 20t cm ( b )At the maximum stretched position, the mass isat the extreme right. Initial phase= π 2 Displacementisgivenas: x=Asin( ωt+ π 2 ) =2sin( 20t+ π 2 ) =2cos 20t ( c )At the maximum compressed position, the mass is at the extreme left. Initial phase= 3π 2 Displacementisgivenas: x=Asin( ωt+ 3π 2 ) =2sin( 20t+ 3π 2 ) =2cos 20t The functions have the same frequency( 20 2π ) andamplitude(2cm), but theydifferin initial phase. 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Q.11 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.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Ans.

( a )Here,time periodofrevolution,T=2s Amplitude,A=3 cm At time,t=0,the radius vector OP makes an angle π 2 withthe positivedirectionofxaxis. Phase angle,ϕ=+ π 2 The equation of SHM for thexprojection of OP,at timet, is given as: x =Acos[ 2πt T +ϕ ] =3cos[ 2πt 2 + π 2 ]=3sin( 2πt 2 ) x=3sinπtcm ( b )Here,time periodofoscillation,T=4 s Amplitudeofoscillation,a=2 m At timet=0, theradiusvectorOP makes an angleπ with thexaxis, in the anticlockwise direction. Phase angle,ϕ=+ π The equation of SHM for thexprojection of theradiusvectorOP, at timet, is given bytherelation: x=acos[ 2πt T +ϕ ]=2cos[ 2πt 4 +π ] x=2cos( πt 2 )m, is the required equation of SHM 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Q.12 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).
(a) x = –2 sin (3t + p /3)
(b) x = cos (p/6 – t)
(c) x = 3 sin (2pt + p/4)
(d) x = 2 cos pt

Ans.

StandardSHM equation isgivenas: x=Acos( 2π T t+ϕ ) ( a )Here,x=2sin( 3t+ π 3 )=+2cos( 3t+ π 3 + π 2 ) =2cos( 3t+ 5π 6 ) Comparingtheabove equation with the standardSHM equation,weobtain: Amplitude,A=2cm Phaseangle,ϕ= 5π 6 = 150 0 Angularvelocity,ω= 2π T =3rad sec 1 The motion of the particle can be representedasshown in the given figure. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@91C9@

( b )Here,x=cos( π 6 t )=cos( t π 6 ) Comparing theabove equation with the standard SHMequation,we obtain: Amplitude,A=1cm Phaseangle,ϕ= π 6 = 30 0 Angularvelocityisgivenas: ω= 2π T =1rad sec 1 The motion of the particle canberepresentedasshown in thegiven figure. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@4C15@

( c )Here,x=3sin( 2πt+ π 4 ) =3cos[ ( 2πt+ π 4 )+ π 2 ]=3cos( 2πt π 4 ) Comparingtheabove equation with the standardSHMequation ,we obtain: Amplitude,A=3 cm Phase angle,ϕ= π 4 =45° Angular velocity,ω= 2π T =2πrad s 1 The motion of the particle can be represented as shown in the given figure. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@5B2C@

( d )Here,x=2 cos πt If wecomparetheabove equation with the standardSHM equation ,we have: Amplitude,A=2 cm Phase angle, ϕ=0 Angular velocity, ω=π rad s -1 Motion of the particle can be represented as shownin the given figure. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1019@

Q.13 Figure 14.26 (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 14.26 (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. 14.26(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) are released, what is the period of oscillation in each case?

Ans.

( a )Incaseof the one block system: When a force( F ) is exerted to the free end of thespring, an extension( l ) is developed. For the maximum valueofextension, wehave: F=kl Here,k= Spring constant The maximum extension in the spring: l= F k Incaseoftwo block system: Here, displacement,x= l 2 Total force,F=+2kx=2k l 2 l= F k ( b )Incaseof one block system: Let mass of the block=m Forceisgiveas:F=ma=m d 2 x d t 2 Here,x=Displacement of the block in timet m d 2 x d t 2 =kx Itsvaue is negativeas the direction of elastic force isopposite to the direction ofdisplacement. d 2 x d t 2 =( k m )x= ω 2 x Here,angular frequency,ω= k m Time periodisgivenas: T= 2π ω =2π m k For two block system: Force,F=m d 2 x d t 2 When the two masses are released, effective mass, m’ = m 1 x m 2 m 1 + m 2 = m 2 m 2 d 2 x d t 2 =kx Itsvalue is negative as the direction of elastic force isopposite to the direction ofdisplacement. d 2 x d t 2 =[ 2k m ]x= ω 2 x Here,angularfrequency,ω= 2k m Timeperiodisgivenas:T= 2π ω =2π m 2k 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Q.14 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?

Ans.

Here,strokeofpiston=1.0 m Angular frequency of piston, ω=200 rad min -1 Amplitude,A= 1.0 2 =0.5m The maximum speed ( v max ) of the piston is given as: v max =Aω=200×0.5=100m min 1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@EB7F@

Q.15 The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)

Ans.

Here,acceleration due to gravity on the moon’ssurface,g’ = 1 .7 ms –2 Acceleration due to gravity on the earth’ssurface,g = 9 .8 ms –2 Time period of a simple pendulum on earth, T e = 3.5 s As T e = 2π l g Here,l= Length of pendulum l = T e 2 2 ×g = 3.5 s 2 3.14 2 ×9.8 ms –2 The valueofthe length of the pendulumremains constant. On the surfaceofmoon, time periodisgivenas: T’ = 2π l g’ = 2π 3.5 s 2 3.14 2 ×9 .8 ms –2 1 .7 ms –2 = 8.4s Time period of the simple pendulum on the surface of moon = 8.4 s

Q.16

Answer the following questions: ( a ) Time period of a particle in SHM depends onthe force constantkand mass m of the particle: T=2π m k .A simple pendulum executes SHMapproximately.Why then is the time period of a pendulum independent of the mass of thependulum? ( b ) The motion of a simple pendulum isapproximately simple harmonic for smallangle oscillations. For larger angles of oscillation, amore involved analysis shows thatTis greater than 2π l g . Think of a qualitative argument toappreciate this result. ( c ) A man with a wristwatch on his hand fallsfrom the top of a tower. Does the watch give correcttime during the free fall? ( d ) What is the frequency of oscillation of asimple pendulum mounted in a cabin that is freely falling under gravity? 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Ans.

( a )The time period of a simple pendulumisgivenas: T=2π m k Incaseof a simple pendulum, km m k =Constant The time period of a simple pendulum doesnotdependonthe mass of thebob. ( b )Fora simple pendulum, the restoring force actingon the bob of thependulum is given bytherelation: F=mgsinθ For smallvaluesofθ, sinθθ For largevaluesofθ, sinθ>θ It decreases the effective value ofaccelerationduetogravityg. The time period increases as: T=2π l g Here,l= Length of the simple pendulum ( c )Yes,becausetheworkingofawristwatchis notbased on pendulum movement, rather, it works on spring action. Hence, itsworkingisnot affected bythe acceleration due to gravity duringfree fall. (d)Whenasimplependulumfallsfreelyundergravity,itsaccelerationiszero.Therefore,thefrequencyof vibrationofthissimplependulumiszero. 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Q.17 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?

Ans.

Here,the bob of the simple pendulum will experiencethe acceleration due to gravity andthe circular motion of the carwillprovidethecentipetalacceleration. Acceleration due to gravity=g Centripetal accelerationisgivenas: v 2 R Here,v=Uniform speed of the car R=Radius of the circulartrack Effective acceleration ofthependulum( a eff ) is givenbytherelation: a eff = g 2 + ( v 2 R ) 2 Time periodofthependulumisgivenas: T=2π l a eff Here,l=Length of the pendulum Time periodofthependulum,T=2π l g 2 + v 4 R 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1EE8@

Q.18

A cylindrical piece of cork of density of base areaAand heighthfloats in a liquid of density. The cork is depressed slightly and then released. Show thatthe cork oscillates up and down simple harmonically with a period T=2π hρ ρ l g whereρis the density of cork. (Ignore dampingdue to viscosity of the liquid). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@48EA@

Ans.

Here,base area of cork=A Height of cork=h Density of liquid= ρ l Density of cork=ρ At equilibrium: Weight of cork=Weight of liquid displaced byfloating cork Let the cork is depressed slightly byxandreleased. Asaresultofit,some extra water of a definite volumeis displaced. An extra upthrust acts intheupwarddirection andprovidestherestoring force to the cork. Upthrust = Restoring force LetF= Weight of extra water displaced F= ( Volume × Density ×g ) Volume = Area × Distance by which the cork is depressed Volume =Ax F= A x ρ l g ( i ) Asper the force law: F=kx Here,kis a constant -k= F x =A ρ l g(ii) Time period of the oscillations of corkisgivenas: T=2π m k (iii) Here,m= mass of cork =volumeofcork×density = Base area of cork × Height of cork × Density of cork=Ahρ Time period ofoscillationofcork: T=2π Ahρ A ρ l g =2π hρ ρ l g MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWabeqaeaaakqaabeqaaiaabIeacaqGLbGaaeOCaiaabwgacaqGSaGaaGjbVlaabkgacaqGHbGaae4CaiaabwgacaqGGaGaaeyyaiaabkhacaqGLbGaaeyyaiaabccacaqGVbGaaeOzaiaabccacaqGJbGaae4BaiaabkhacaqGRbGaaGjbVlabg2da9iaaysW7caqGbbaabaGaaeisaiaabwgacaqGPbGaae4zaiaabIgacaqG0bGaaeiiaiaab+gacaqGMbGaaeiiaiaabogacaqGVbGaaeOCaiaabUgacaaMe8Uaeyypa0JaaGjbVlaabIgaaeaacaqGebGaaeyzaiaab6gacaqGZbGaaeyAaiaabshacaqG5bGaaeiiaiaab+gacaqGMbGaaeiiaiaabYgacaqGPbGaaeyCaiaabwhacaqGPbGaaeizaiaaysW7cqGH9aqpcaaMe8UaeqyWdi3aaSbaaSqaaiaadYgaaeqaaaGcbaGaaeiraiaabwgacaqGUbGaae4CaiaabMgacaqG0bGaaeyEaiaabccacaqGVbGaaeOzaiaabccacaqGJbGaae4BaiaabkhacaqGRbGaaGjbVlabg2da9iaaysW7cqaHbpGCaeaacaqGbbGaaeiDaiaabccacaqGLbGaaeyCaiaabwhacaqGPbGaaeiBaiaabMgacaqGIbGaaeOCaiaabMgacaqG1bGaaeyBaiaabQdaaeaacaqGxbGaaeyzaiaabMgacaqGNbGaaeiAaiaabshacaqGGaGaae4BaiaabAgacaqGGaGaae4yaiaab+gacaqGYbGaae4AaiaaysW7caqG9aGaaGjbVlaabEfacaqGLbGaaeyAaiaabEgacaqGObGaaeiDaiaabccacaqGVbGaaeOzaiaabccacaqGSbGaaeyAaiaabghacaqG1bGaaeyAaiaabsgacaqGGaGaaeizaiaabMgacaqGZbGaaeiCaiaabYgacaqGHbGaae4yaiaabwgacaqGKbGaaeiiaiaabkgacaqG5bGaaGjbVlaabAgacaqGSbGaae4BaiaabggacaqG0bGaaeyAaiaab6gacaqGNbGaaeiiaiaabogacaqGVbGaaeOCaiaabUgaaeaacaqGmbGaaeyzaiaabshacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqGVbGaaeOCaiaabUgacaqGGaGaaeyAaiaabohacaqGGaGaaeizaiaabwgacaqGWbGaaeOCaiaabwgacaqGZbGaae4CaiaabwgacaqGKbGaaeiiaiaabohacaqGSbGaaeyAaiaabEgacaqGObGaaeiDaiaabYgacaqG5bGaaeiiaiaabkgacaqG5bGaaGjbVlaabIhacaaMe8Uaaeyyaiaab6gacaqGKbGaaGjbVlaabkhacaqGLbGaaeiBaiaabwgacaqGHbGaae4CaiaabwgacaqGKbGaaeOlaiaabccaaeaacaqGbbGaae4CaiaaysW7caqGHbGaaGjbVlaabkhacaqGLbGaae4CaiaabwhacaqGSbGaaeiDaiaaysW7caqGVbGaaeOzaiaaysW7caqGPbGaaeiDaiaabYcacaaMe8Uaae4Caiaab+gacaqGTbGaaeyzaiaabccacaqGLbGaaeiEaiaabshacaqGYbGaaeyyaiaabccacaqG3bGaaeyyaiaabshacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeiiaiaabsgacaqGLbGaaeOzaiaabMgacaqGUbGaaeyAaiaabshacaqGLbGaaeiiaiaabAhacaqGVbGaaeiBaiaabwhacaqGTbGaaeyzaiaaysW7caqGPbGaae4CaiaabccacaqGKbGaaeyAaiaabohacaqGWbGaaeiBaiaabggacaqGJbGaaeyzaiaabsgacaqGUaaabaGaeyinIWLaaeyqaiaab6gacaqGGaGaaeyzaiaabIhacaqG0bGaaeOCaiaabggacaqGGaGaaeyDaiaabchacqGHsislcaqG0bGaaeiAaiaabkhacaqG1bGaae4CaiaabshacaqGGaGaaeyyaiaabogacaqG0bGaae4CaiaabccacaqGPbGaaeOBaiaaysW7caqG0bGaaeiAaiaabwgacaaMe8UaaeyDaiaabchacaqG3bGaaeyyaiaabkhacaqGKbGaaGjbVlaabsgacaqGPbGaaeOCaiaabwgacaqGJbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGHbGaaeOBaiaabsgacaaMe8UaaeiCaiaabkhacaqGVbGaaeODaiaabMgacaqGKbGaaeyzaiaabohacaaMe8UaaeiDaiaabIgacaqGLbGaaGjbVlaabkhacaqGLbGaae4CaiaabshacaqGVbGaaeOCaiaabMgacaqGUbGaae4zaiaabccacaqGMbGaae4BaiaabkhacaqGJbGaaeyzaiaabccacaqG0bGaae4BaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaae4yaiaab+gacaqGYbGaae4Aaiaab6caaeaacaqGvbGaaeiCaiabgkHiTiaabshacaqGObGaaeOCaiaabwhacaqGZbGaaeiDaiaabccacqGH9aqpcaqGGaGaaeOuaiaabwgacaqGZbGaaeiDaiaab+gacaqGYbGaaeyAaiaab6gacaqGNbGaaeiiaiaabAgacaqGVbGaaeOCaiaabogacaqGLbaabaGaaeitaiaabwgacaqG0bGaaGjbVlaabAeacqGH9aqpcaqGGaGaae4vaiaabwgacaqGPbGaae4zaiaabIgacaqG0bGaaeiiaiaab+gacaqGMbGaaeiiaiaabwgacaqG4bGaaeiDaiaabkhacaqGHbGaaeiiaiaabEhacaqGHbGaaeiDaiaabwgacaqGYbGaaeiiaiaabsgacaqGPbGaae4CaiaabchacaqGSbGaaeyyaiaabogacaqGLbGaaeizaaqaaiaabAeacqGH9aqpcaqGGaGaai4eGiaabccadaqadaqaaiaabAfacaqGVbGaaeiBaiaabwhacaqGTbGaaeyzaiaabccacqGHxdaTcaqGGaGaaeiraiaabwgacaqGUbGaae4CaiaabMgacaqG0bGaaeyEaiaabccacqGHxdaTcaqGNbaacaGLOaGaayzkaaaabaGaaeOvaiaab+gacaqGSbGaaeyDaiaab2gacaqGLbGaaeiiaiaab2dacaqGGaGaaeyqaiaabkhacaqGLbGaaeyyaiaabccacqGHxdaTcaqGGaGaaeiraiaabMgacaqGZbGaaeiDaiaabggacaqGUbGaae4yaiaabwgacaqGGaGaaeOyaiaabMhacaqGGaGaae4DaiaabIgacaqGPbGaae4yaiaabIgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaabogacaqGVbGaaeOCaiaabUgacaqGGaGaaeyAaiaabohacaqGGaGaaeizaiaabwgacaqGWbGaaeOCaiaabwgacaqGZbGaae4CaiaabwgacaqGKbaabaGaaeOvaiaab+gacaqGSbGaaeyDaiaab2gacaqGLbGaaeiiaiabg2da9iaabgeacaqG4baabaGaeyinIWLaaeOraiabg2da9iaabccacaGGtaIaaeyqaiaabccacaqG4bGaaGjbVlabeg8aYnaaBaaaleaacaWGSbaabeaakiaaysW7caWGNbGaeyOKH4QaaeiiamaabmaabaGaaeyAaaGaayjkaiaawMcaaaqaaiaabgeacaqGZbGaaGjbVlaabchacaqGLbGaaeOCaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeOzaiaab+gacaqGYbGaae4yaiaabwgacaqGGaGaaeiBaiaabggacaqG3bGaaeOoaaqaaiaabAeacqGH9aqpcqGHsislcaqGRbGaaeiEaaqaaiaabIeacaqGLbGaaeOCaiaabwgacaqGSaGaae4AaiaaysW7caqGPbGaae4CaiaabccacaqGHbGaaeiiaiaabogacaqGVbGaaeOBaiaabohacaqG0bGaaeyyaiaab6gacaqG0baabaGaaeylaiaabUgacaaMe8UaaeypaiaaysW7daWcaaqaaiaadAeaaeaacaWG4baaaiabg2da9iabgkHiTiaabgeacaqGGaGaeqyWdi3aaSbaaSqaaiaadYgaaeqaaOGaaGjbVlaadEgacqGHsgIRcaqGOaGaaeyAaiaabMgacaqGPaaabaGaaeivaiaabMgacaqGTbGaaeyzaiaabccacaqGWbGaaeyzaiaabkhacaqGPbGaae4BaiaabsgacaqGGaGaae4BaiaabAgacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiaiaab+gacaqGZbGaae4yaiaabMgacaqGSbGaaeiBaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqGJbGaae4BaiaabkhacaqGRbGaaGjbVlaabMgacaqGZbGaaGjbVlaabEgacaqGPbGaaeODaiaabwgacaqGUbGaaGjbVlaabggacaqGZbGaaeOoaaqaaiaadsfacqGH9aqpcaaIYaGaeqiWda3aaOaaaeaadaWcaaqa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Q.19 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.

Ans.

Here, crosssectional area of the Utube=A Density of mercury column=ρ Acceleration due to gravity=g Letheight of the mercury column in the two arms=2h Restoring force,F=Weight of the mercury column of a definite height F= ( Volume × Density ×g ) F= ( A× 2h×ρ ×g )=2Aρgh F= k× Displacement in one of the arms ( h ) Here,k=Constant k=- F h = 2Aρgh h =2Aρg Time periodisgivenbytherelation: T=2π m k =2π m 2Aρg Here,m=Mass of the mercury column Lettotal length of themercury column=l Mass of mercury,m= Volume × Density =Alρ Timeperiod,T=2π Alρ 2Aρg =2π l 2g The mercury column undergoes SHM with timeperiod 2π l 2g . 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Q.20 An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.14.27). 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 Fig. 14.27]

Ans.

Here,volume of air chamber=V Crosssectionalarea of the neck=a Letmass of ball=m Pressure inside the chamber=Atmospheric pressure Let the ball be pressedthroughasmalldistancexverticallydownward. As a result, there will be a decrease in the volume and an increase in the pressure inside the chamber. Decrease in the volume ofair chamber, ΔV=ax Volumetric strain= Changeinvolume Originalvolume = ΔV V = ax V Bulk Modulus of airisgivenas: B= Stress Strain = p ax V Here, stress is the increase in pressure. The negativesign shows that thepressure increases with decrease in volume. p= Bax V The restoring force acting on the ballisgivenas: F=p × a = Bax V ×a= B a 2 x V (i) For SHM, the equation for restoringforce isgivenas: F= kx ( ii ) Here,k= Spring constant Comparing equations ( i ) and ( ii ), we obtain: k= B a 2 V (iii) Time periodisgivenbytherelation: T=2π m k (iv) Substitutingthevalueofkfromequation(iii)inequation(iv),weobtain: Timeperiod,T=2π Vm B a 2 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Q.21 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

Ans.

a Here,mass of automobile,m = 3000 kg Displacement in the suspension system,x = 15 cm = 0.15 m Letspringconstantofeachspring be k Asthere are 4 springs in parallel to support the wholemass of the automobile, Springconstantof4 springs,K = 4k The equation for the restoring force for the systemisgivenas: F = –4kx = mg ork = mg 4x = 3000 kg×10 ms -2 4×0.15 m = 5×10 4 Nm -1 Spring constant, k = 5 × 10 4 Nm -1 b Here,masssupportedbyeach wheel,M = 3000 kg 4 = 750 kg For damping factor b, the equation for displacement, x = x 0 e -bt/2M Since amplitude of oscillation decreases by 50%, x = x 0 2 x 0 2 = x 0 e -bt/2M log e 2 = bt 2M b = 2Mlog e 2 t Timeperiod, t = 2π m 4k = 2π 3000kg 4×5×10 4 Nm -1 = 0.7691s b = 2×750 kg×0.693 0.7691 s = 1351 .58 kgs -1 Damping constant of the spring = 1351 .58 kgs -1

Q.22 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.

Ans.

ConsideraparticleofmassmbeexecutingSHMwith timeperiodT. The displacement of the particle executing SHM at an instanttis given bytherelation: x = Asinωt Here,A = Amplitude ω = Angular frequency= k M Velocity of the particle, v = dx dt = Aωcosωt Kinetic energy of the particle, E k = 1 2 Mv 2 = 1 2 MA 2 ω 2 cos 2 ωt Potential energy of the particle, E p = 1 2 kx 2 = 1 2 2 A 2 sin 2 ωt Qk=Mω 2 Lettimeperiod = T Average kinetic energy over one cycle, E k av = 1 T 0 T E k dt = 1 T 0 T 1 2 MA 2 ω 2 cos 2 ωtdt E k av = 1 2T MA 2 ω 2 0 T 1+cos2ωt 2 dt = 1 4T MA 2 ω 2 t+ sin2ωt 0 T E k av = 1 4T MA 2 ω 2 T = 1 4 MA 2 ω 2 (i) Average potential energy over one cycle, E p av = 1 T 0 T E p dt = 1 T 0 T 1 2 2 A 2 sin 2 ωtdt E p av = 1 2T 2 A 2 0 T 1-cos2ωt 2 dt = 1 4T 2 A 2 t- sin2ωt 0 T E p av = 1 4T 2 A 2 T = 1 4 2 A 2 (ii) From equations i and ii itcanbeconcludedthat theaverage kinetic energy for agiven time period is equal to the average potential energy for the same time period.

Q.23 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 = –αθ , where J is the restoring couple and θ the angle of twist).

Ans.

Here,mass of the disc,m = 10 kg Radius of disc,r = 15 cm = 0.15 m Time periodoftorsional oscillations,T = 1.5 s Moment of inertia of the disc, I = 1 2 mr 2 I = 1 2 × 10 kg × 0.15 m 2 = 0 .1125 kgm 2 Time period,T = 2π I α Here,α is the torsional constant. α = 2 I T 2 = 4×π 2 ×0 .1125 kgm 2 1.5 s 2 = 1 .972 Nmrad -1 Torsional spring constant of the wire = 1 .972 Nmrad –1 .

Q.24 A body describes simple harmonic motion with 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.

Ans.

Here,amplitude,A = 5 cm = 0.05 m Time period,T = 0.2 s (a) Here, displacement,x = 5 cm = 0.05 m Acceleration, a = –ω 2 x = – T 2 x = – 0.2 2 ×0 .05 m = -5π 2 ms -2 Velocity, v = ω A 2 – x 2 = T 0.05 m 2 0.05 m 2 v = 0 When the displacement of the body is 5 cm: Acceleration ofthebody = –5π 2 ms -2 Velocity ofthebody=0 (b) Here, displacementofbody, x = 3 cm = 0.03 m Acceleration, a = –ω 2 x = – T 2 x = – 0.2 2 ×0 .03 m = -3π 2 ms -2 Velocity, v = ω A 2 – x 2 = T 0.05 m 2 0.03 m 2 v = 0.2 s ×0.04 m = 0 .4π ms -1 When the displacement of the body is 3 cm: Acceleration ofthebody, a = –3π 2 ms -2 Velocity ofthebody,v = 0 .4π ms -1 (c)Here, displacementofbody,x = 0 Acceleration, a = -ω 2 x = 0 Velocity, v = ω A 2 – x 2 = T 0.05 m 2 – (0 m) 2 = 0.5π ms -1 When the displacement of the body is0: Acceleration ofthebody =0 Velocity ofthebody=0.5π ms -1

Q.25 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 x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation x = a cos (ωt + θ) and note that the initial velocity is negative.]

Ans.

The displacement of an oscillating mass is given bytherelation: x=Acos( ωt+θ ) Here,A=Amplitudeofoscillations x= Displacementofthemass θ=Phase constant Velocityisgivenas:v= dx dt =Aωsin( ωt+θ ) Attime,t=0,x= x 0 x 0 =Acosθ(i) dx dt = v 0 =Aωsinθ Asinθ= v 0 ω ( ii ) Squaring and adding equations ( i ) and ( ii ), we obtain: A 2 ( cos 2 θ+ sin 2 θ )= x 0 2 + ( v 0 ω ) 2 A= x 0 2 + ( v 0 ω ) 2 Amplitude of the resulting oscillation= x 0 2 + ( v 0 ω ) 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D2B5@

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FAQs (Frequently Asked Questions)

1. Where can I get the answers for NCERT Solutions for Class 11 Physics Chapter 14?

 On Extramarks website, you can find the correct NCERT Solutions for Class 11 Physics Chapter 14. Physics  faculty at Extramarks have meticulously created the NCERT Textbook Solutions. All of these solutions are based on the CBSE’s new term-by-term test format, so that students can prepare thoroughly for their second-term exams.

2. How to Use Class 11 chapter 4 Oscillations Solutions?

Once you’ve accessed Extramarks NCERT solutions, you can put them to good use to learn and improve your grades and in case you are unable to solve a question, review the themes and solutions. While revising, use these as a reference guide.

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Extramarks NCERT Solutions for Class 11 Physics Chapter 14 Oscillations have a number of outstanding features.  Some of them are as follows:

  • It’s a fantastic tool for studying and preparing for exams.
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4. Define oscillation.

Any periodic motion of an object at a distance around the equilibrium position that repeats itself over a length of time is characterized as oscillation. Oscillation up and down of a spring, oscillation swinging side by side of a pendulum also known as Simple harmonic motion and so on are examples of oscillation.

5. Explain forced oscillation.

The oscillation where the body oscillates under the influence of an external periodic force is known as forced oscillation. When learning about forced oscillations, it’s vital to remember the phrase resonance. The frequency of the external force must be equal to the oscillator’s natural frequency to be considered resonant. The resonant frequency is another name for this frequency.

6. How to prepare for Class 11 Physics Chapter 14?

To do so,  students must read the chapter thoroughly and highlight all  key points that they consider vital.  Besides this, the different formulas and laws endorsed in this chapter are  crucial for the learner to be familiar with and practice all questions to test their understanding and for self-assessment. They should also refer to the NCERT solutions to help them  brush up  their concepts. These basic measures will assist the learner in performing well on the exam.

7. What is the Doppler effect?

According to the Doppler effect, when there is a relative motion between the source of sound and the listener, the frequency of the sound heard by the listener is completely different from the frequency of the sound emitted by the source.

8. What are the topics that are covered in Class 11 Physics Chapter 14?

The following topics have been covered  in Class 11 Physics Chapter 14: 

  • Introduction
  • Periodic and oscillatory motion
  • Simple harmonic motion
  • Simple harmonic motion and uniform circular motion
  • Velocity and acceleration in simple harmonic motion
  • Force law for simple harmonic motion
  • Energy in simple harmonic motion
  • Some systems execute simple harmonic motion
  • Damped simple harmonic motion
  • Forced oscillation and resonance

Students should refer to Extramarks NCERT solutions for further information on this chapter and practice different types of question papers, MCQs, sample papers etc. . These exercises are meant to cover the key points from this chapter, allowing the student to gain a solid understanding of the chapter and thus perform well on the exam. Extramarks makes all of the study materials available for free.

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