NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations
Have you ever observed the action of a swing going back and forth, of a pendulum going back and forth, or the action of springs and musical instruments? All these phenomena in our daily life can be attributed to the principles of oscillatory motion, which are the pillars of NCERT Class 11 Physics Chapter 14 Oscillations. This chapter enables students to have an idea about periodic motion, restoring forces, and the change of energy in oscillating systems.
This Story also Contains
- NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQI
- NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQII
- NCERT Exemplar Class 11 Physics Solutions Chapter 14: Very Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 14: Short Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 14: Long Answer
- NCERT Exemplar Class 11 Physics Solutions Chapter 14 – Oscillations: Important Concepts and Formulas
- Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations
- Approach to Solve Exemplar Questions of Chapter 14 Oscillations
- NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
- NCERT Exemplar Solutions Class 11 Subject-wise
- NCERT Solutions for Class 11 Physics Chapter-wise
In NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations, concepts, which include simple harmonic motion (SHM), displacement, velocity and acceleration in SHM, time period, frequency, phase, energy in oscillatory motion and oscillations of spring-mass and pendulum systems, are of significance. The concepts form a fundamental aspect of explaining real-life vibrations and are common in CBSE board exams, not to mention competitive exams like JEE. The NCERT Exemplar solutions are given in a step-wise, straightforward manner, such that numerical problems and theory-based queries are not hard to understand and can be solved easily. Prepared by experienced subject matter experts, NCERT Exemplar Solutions Class 11 Physics Chapter 14 strictly follow the latest CBSE syllabus and NCERT guidelines. All solutions are aimed at enhancing their conceptual understanding, minimising typical errors, and establishing a logical approach to overcoming the problems of oscillation. These questions are considered to be good to practice regularly to improve the ability to think analytically and the confidence in solving problems. It has a great emphasis on the practical aspects of life and methodical explanations, and NCERT Exemplar Class 11 Physics Solutions on Oscillations would be a great study resource to revise and prepare to an exam.
Also Read
- Oscillations Class 11th Notes - Free NCERT Class 11 Physics Chapter 14 Notes Download PDF
- NCERT Solutions for Class 11 Physics Chapter 14 Oscillations
NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQI
Exemplar Class 11 Physics Solutions Chapter 14 Oscillations: MCQ I are used to test the simple harmonic motion and oscillatory motion basic knowledge of the students using objective questions that require only one correct choice. Such MCQs assist in enhancing conceptual understanding, precision, and rapid recall, thus they are very useful in rapid revision and exam preparation.
Question:14.1
The displacement of a particle is represented by the equation $y = 3 \cos\left (\frac{\pi}{4} - 2 \omega t \right )$. The motion of the particle is
(a) simple harmonic with period $\frac{2p}{w}$
(b) simple harmonic with period $\frac{\pi}{\omega}$
(c) periodic but not simple harmonic
(d) non-periodic
Answer:
The answer is option (b), Simple harmonic with period $\frac{\pi}{\omega}$Question:14.2
The displacement of a particle is represented by the equation $y= \sin^{3} \omega t$. The motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period $\frac{2\pi }{\omega}$
(d) simple harmonic with period $\frac{\pi }{\omega}$
Answer:
The answer is option (b), periodic but not simple harmonicQuestion:14.3
The relation between acceleration and displacement of four particles are given below:
(a) $a_{x} = +2x$
(b) $a_{x} = +2x^{2}$
(c) $a_{x} = -2x^{2}$
(d) $a_x = -2x$
Which one of the particle is exempting simple harmonic motion?
Answer:
The answer is option (d) $a_x = -2x$Explanation: In simple harmonic motion,
Acceleration is proportional (as well as opposite) to displacement.
Thus, opt (d)
Question:14.4
The motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the density of the liquid
(d) simple harmonic and time period is directly proportional to the density of the liquid
Answer:
The answer is the option (c) Simple harmonic and time period is independent of the density of the liquid.Explanation: Let us take a test tube viz., filled with a liquid of density $\rho$ up to height ‘h’.
When the liquid is lifted in arm Q to a height ‘y’ from A to B, the liquid in arm P drops by the height ‘y’ from A’ to C’. The height difference between the two arms is 2y.
Here, the hydrostatic pressure provides the restoring force, thus,
$F = -V. \rho.g$
$= -A.2y\rho g$ , where,
A = Area of a cross-section of tube, & $F \propto -y,$
Thus, it is simple harmonic motion.
$T = \sqrt{\frac{2\pi m (inertia)}{k (spring)}}= \sqrt{\frac{2\pi A (2h)}{\rho 2 A \rho g}}$
Thus, $T = 2\pi \sqrt{\frac{h}{g}}$
Thus, the motion is harmonic as the time period is independent of density.
Question:14.5
A particle is acted simultaneously by mutually perpendicular simple harmonic motion $x = a \cos \omega t and y = a \sin \omega t$. The trajectory of motion of the particle will be
(a) an ellipse
(b) a parabola
(c) a circle
(d) a straight line
Answer:
The answer is the option (c) a circle.Explanation: We know that,
Resultant displacement = x + y
$= a \cos \omega t + a \sin \omega t$
Thus, $y^{'} = a \left (\cos \omega t + a \sin \omega t \right )$
$=a \sqrt{2}\left [ \frac{\cos \omega t}{\sqrt{2}}+ \frac {\sin \omega t}{\sqrt{2}} \right ]$
$=a \sqrt{2} \left [ \cos \omega t \cos 45^{\circ} +\sin \omega t \sin 45^{\circ} \right ]$
……. (particle is acted simultaneously by mutually perpendicular directions)
$y^{'}= a\sqrt{2}\cos s\left ( \omega t - 45^{\circ} \right )$ ……. Thus, the displacement can neither be a straight line nor a parabola
Now, let us square and add x & y,
$x^{2} + y^{2} = a^{2} \cos^{2} \omega t + a^{2} \sin^{2} \omega t$
Thus, $x^{2} + y^{2} = a^{2}$
This is the equation of a circle, hence, opt (c)
Question:14.6
The displacement of a particle varies with time according to the relation
$y = a \sin t + b \cos t$
(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude $a^{2} + b^{2}$
(d) The motion is SHM with amplitude $\sqrt{{a}^{2} + b^{2}}$
Answer:
The answer is the option (d) The motion is SHM with amplitude $\sqrt{a^{2}+b^{2}}$Explanation: Given: $y = a \sin \omega t + b \cos \omega t$
Now, we know that,
The amplitude of motion, $A = \sqrt{a^{2}+b^{2}}$
Thus, opt (d).
Question:14.7
Four pendulums A, B, C and D are suspended from the same elastic support as shown in the figure. A and C are of the same length, while B is smaller than A and D is larger than A. If A is given a transverse displacement,
(a) D will vibrate with maximum amplitude
(b) C will vibrate with maximum amplitude
(c) B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude.
Answer:
The answer is the option (b) C will vibrate with maximum amplitude.Explanation: If the pendulum vibrates with transverse vibration,
Time period, $T=2\pi \sqrt{\frac{l}{g}}$, where,
l = length of pendulum A & C.
Now, the disturbance produced is transmitted to all pendulums, i.e., B, C & D, where the time period (T) of C & A is the same. C will vibrate with a maximum in resonance, as the periodic force of period T produces resonance in C.
Hence, opt (b).
Question:14.8
Figure shows the circular motion of a particle. The radius of the circle, the period, sense of revolution and the initial position are indicated on the figure. The simple harmonic motion of the x-projection of the radius vector of the rotating particle P is
$(a)x(t)= B\sin \left ( \frac{2\pi t}{30} \right )$
$(b)x(t)= B\sin \left ( \frac{\pi t}{15} \right )$
$(c)x(t)= B\sin \left ( \frac{\pi t}{15}+\frac{\pi}{2} \right )$
$(d)x(t)= B\cos \left ( \frac{\pi t}{15}+\frac{\pi}{2} \right )$
Answer:
The answer is the option $(a)x(t)= B\sin \left ( \frac{2\pi t}{30} \right )$Explanation: Circular motion is executed by P with radius B.
Here, $x = OQ \cos (90^{\circ} - \theta)$
=$OQ \sin \theta$
= $B \sin \omega$
Thus, $\theta = \omega t$
$x = B \sin \frac{2 \pi }{T} t$
$x = B \sin \frac{2 \pi }{30}t$
Thus, opt (a)
Question:14.9
The equation of motion of a particle is $x = a \cos (\alpha t)^{2}.$
The motion is
(a) periodic but not oscillatory.
(b) periodic and oscillatory.
(c) oscillatory but not periodic.
(d) neither periodic nor oscillatory.
Answer:
The answer is the option (c) Oscillatory but not periodic.Question:14.10
A particle executing S.H.M. has a maximum speed of $30 \frac{cm}{s}$ and a maximum acceleration of $60 \frac{cm}{s^{2}}$. The period of oscillation is
$(a) \pi s$.
$(b) \left ( \frac{\pi}{2} \right ) s$.
$(c) 2\pi s$.
$(d) \left (\frac{\pi}{t} \right ) s.$
Answer:
Answer: The answer is the option $(a) \pi s$Explanation: Let us consider the equation of a SHM
Thus,$y = a \sin \omega t$
$v = \frac{dy}{dt}$
$= a\omega \cos \omega t$ ……. (i)
$a = \frac{dv}{dt}$
$= -a\omega^ 2 \sin \omega t$ ….. (ii)
$V_{max} = 30 \frac{cm}{s}$ …. (given)
From (i),
$V_{max} = a \omega$
Thus, $a \omega=30$ ….. (iii)
From (ii),
Now, $a_{max} = a\omega ^{2} = 60$ …….. (iv)
Thus, $60 = \omega \times 30$
Therefore, $\omega = 2 \frac{rad}{s}$
Now, $\frac{2 \pi }{T} = 2$
Thus, $T = \pi$.
Hence opt (a).
Question:14.11
Answer:
Answer: The answer is the option $(b) \sqrt{v_{1}^{2}+v_{2}^{2}}$Explanation: If we connect a mass (m) to a spring on a frictionless horizontal surface, then their frequencies will be –
$V_{1} = \frac{1}{2\pi}\sqrt{\frac{K_{1}}{m}}$ & $V_{2} = \frac{1}{2\pi}\sqrt{\frac{K_{2}}{m}}$
…. (i) …. (ii)
Since the springs are parallel, their equivalent will be-
$K_{p} = K_{1} + K_{2}$
& frequency will be –
$V_{p} = V_{1} =\frac{ 1}{2 \pi}\sqrt{\frac{K_{1} + K_{2}}{m}}$
$=\frac{ 1}{2 \pi}\left [ \frac{K_{1}}{m}+\frac{K_{2}}{m} \right ]^{\frac{1}{2}}$
From (i),
$\frac{ K_{1}}{m} = (2\pi v_{1})^{2} = 4\pi^{ 2}v_{1}^{2}$
& $\frac{ K_{2}}{m} = (2\pi v_{1})^{2} = 4\pi^{ 2}v_{2}^{2}$
Thus, $V_{p} = \frac{1}{2\pi} [4\pi ^{2}v_{1}^{2} + 4\pi ^{2}v_{2}^{2}]$
$V_{p} = \sqrt{v_{1}^{2}+v_{2}^{2}}$
Hence, opt (b)
NCERT Exemplar Class 11 Physics Solutions Chapter 14: MCQII
NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations: MCQII centre on multiple-answer and reasoning-based objective tests which examine more in-depth conceptual learning of oscillatory movement and SHM. These questions assist the students to critically examine the conditions, eliminate misunderstandings and develop confidence in managing the advanced MCQs in the exam.
Question:14.12
The rotation of earth about its axis is
(a) periodic motion
(b) simple harmonic motion
(c) periodic but not simple harmonic motion
(d) non-periodic motion
Answer:
The answer is the option (a) periodic motion and (c) periodic but not SHMExplanation:
(i) The Earth completes one revolution in a regular interval of time.
(ii) The motion of the earth is circular about its own axis.
(iii) But this motion is not SHM, as we cannot measure its displacement since it is not about a fixed point. Also, it does not move both sides.
Question:14.13
Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower point is
(a) simple harmonic motion
(b) non-periodic motion
(c) periodic motion
(d) periodic but not SHM
Answer:
The answer is the option (a) Simple harmonic motion and (b) non - periodic motionExplanation: Lift the ball from point A to B and smoothly release it to reach C & then return to point A and then B. Thus, it is a periodic motion. Here, mg sin $\theta$ balances the restoring force (R); also, a restoring force $(mg \sin \theta)$acts on the ball.
Thus, $ma = mg \sin \theta$
$a = g \sin \theta$ or $\frac{d^{2}x}{dt^{2} }= - g \sin \theta$ ….. (when the ball moves upward)
$\frac{d^{2}x}{dt^{2} }= - g \left ( \frac{x}{r} \right )$
Thus, $\frac{d^{2}x}{dt^{2} }\propto (-x)$
Thus, it is a simple harmonic motion.
$\Omega =\sqrt{\frac{ g}{r}}$ or $T = \frac{2\pi }{\omega} = 2\pi \sqrt{\frac{r}{g}}$
Thus, this motion is periodic as well as simple harmonic.
Question:14.14
Displacement versus time curve for a particle executing SHM is shown in figure. Choose the correct statements.
(a) Phase of the oscillator is same at t = 0s and t = 2 s
(b) Phase of the oscillator is same at t = 2s and t = 6 s
(c) Phase of the oscillator is same at t = 1s and t = 7 s
(d) Phase of the oscillator is same at t = 1s and t = 5 s
Answer:
The answer is the option (b) Phase of the oscillator is the same at t = 2s & t = 6s and (d) The phase of the oscillator is the same at t = 1s & t = 5sExplanation: If the mode of vibration of two particles is the same, then they are said to be in the same phase. That means, their distance will be $n\lambda$, where, n = 1, 2, 3,4, …..
(a) Here, the particles are not in the same phase as the distance between them is $\frac{\lambda }{2}$.
(b) The particles are at a distance $\lambda$. Thus, they are in the same phase.
(c) The distance between particles is $\lambda =\frac{\lambda }{2}=\frac{3\lambda }{2}$. Thus, they are not in the same phase.
(d) The distance between the particles is $\lambda$. Thus, they are in the same phase.
Question:14.15
Which of the following statements is/are true for a simple harmonic oscillator?
(a) Force acting is directly proportional to displacement from the mean position and opposite to it
(b) Motion is periodic
(c) Acceleration of the oscillator is constant
(d) The velocity is periodic
Answer:
The answer is the option (a) Force acting is directly proportional to the displacement from the mean position and opposite to it, (b) Motion is periodic, and (d) The velocity is periodic.Explanation: Let, $x = a \sin \omega t$, be a SHM …… (i)
$V = \frac{dx}{dt} = \alpha \omega \cos \omega$ …. (ii)
$V = \frac{dv}{dt} = \alpha \omega\left ( - \omega \sin \omega t \right )$
Thus, $A = - \alpha \omega ^{2} \sin \omega t$ or A = -ω2x
$mA = -m\omega ^{2}x$
Hence, $F \propto (-x)$
Now, x(t) = x(t + T)
Thus, the motion is periodic and simple harmonic.
Now, from (ii),
v(t) = v(t + T)
Hence, opt, (a), (b) & (d) are correct options.
Question:14.16
The displacement time graph of a particle executing S.H.M. is shown in Figure. Which of the following statement is/are true?
(a) The force is zero at $t = \left (\frac{3T}{4} \right )$
(b) The acceleration is maximum at $t = \left (\frac{4T}{4} \right )$
(c) The velocity is maximum at $t = \left (\frac{T}{4} \right )$
(d) The P.E. is equal to K.E. of oscillation at $t = \left (\frac{T}{2} \right )$
Answer:
The answer is the option (a) The force is zero at $t = \left (\frac{3T}{4} \right )$, (b) The acceleration is maximum at $t = \left (\frac{4T}{4} \right )$, and (c) The velocity is maximum at $t = \left (\frac{T}{4} \right )$Explanation: (a) The particle is at its mean position at $t = \left (\frac{3T}{4} \right )$, so the force acting on it is zero, but due to the inertia of mass, the motion continues
a = 0 thus, F = 0.
(b) Particle's velocity changes from increasing to decreasing, with a maximum at the change in velocity, at $t = \left (\frac{4T}{4} \right )$. Thus, acceleration is maximum here.
(c) The velocity is maximum at its mean position at $t = \left (\frac{T}{4} \right )$ as there is no retarding force on it.
(d) K.E. = 0 at $t = \frac{T}{2} = \frac{2T}{4}.$
Thus, P.E is not equal to kinetic energy.
Question:14.17
A body is performing SHM, then its
(a) The average total energy per cycle is equal to its maximum kinetic energy
(b) The average kinetic energy per cycle is equal to half of its maximum kinetic energy
(c) mean velocity over a complete cycle is equal to $\frac{2}{\pi}$ times of its maximum velocity
(d) root mean square velocity is $\frac{1}{\sqrt{2}}$ times of its maximum velocity
Answer:
The answer is the option (a) Average total energy per cycle is equal to its maximum kinetic energy, (b) Average kinetic energy per cycle is equal to half its maximum kinetic energy, and (d) Root square mean velocity is equal to $\frac{1}{\sqrt{2}}$ times its maximum velocity.Explanation: (a) let$x = a \sin \omega t$ be a periodic SHM
Let m be the mass executing SHM
$v = \frac{dx}{dt}$
= $a\omega \cos \omega t$
$v_{max} = \alpha \omega ..... (since \cos \omega t = 1)$
Now, $\frac{K.E.}{P.E}$ = Total mechanical energy
Thus, $T.E. = \frac{1}{2} ma^{2} \omega ^{2}$
Thus, $K.E. = \frac{1}{2} ma^{2} \omega ^{2}$
Or we can say that the average total energy is K.E.max
(b) Let, amplitude = a
Angular frequency = $\omega$
Thus, maximum velocity = αω, which varies according to the sine law.
Thus, the rms value of a complete cycle = $-\frac{1}{\sqrt{2}} \alpha \omega .$
Thus, the average
$K.E. = \frac{1}{2} mv_{rms}^{2}$
= $\frac{1}{2}m \frac{1}{2} a^{2}\omega ^{2}$
= $\frac{1}{2}\left [m \frac{1}{2} a^{2}\omega ^{2} \right ]$
= $\frac{1}{2}\left [ \frac{1}{2} m _{max}^{2} \right ]$
=$\frac{1}{2} K.E. _{max}$
(c) $v = a\omega \cos \omega t$
$V_{mean} = \frac{(a\omega -a\omega )}{2}$
Vmean = 0 ….. since vmax is not equal to vmean
$V_{rms} = \sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{2} }= \frac{a\omega }{\sqrt{2}}$
Thus, $v_{rms }= \frac{v_{max}}{\sqrt{2}}$
Question:14.18
A particle is in linear simple harmonic motion between two points A and B, 10 cm apart (figure). Take the direction from A to B as the positive direction and choose the correct statements. __ _
AO = OB = 5 cm
BC= 8 cm
(a) The sign of velocity, acceleration and force on the particle when it is 3 cm away from A going towards B are positive
(b) The sign of velocity of the particle at C going towards B is negative
(c) The sign of velocity, acceleration and force on the particle when it is 4 cm away from B going towards A are negative
(d) The sign of acceleration and force on the particle when it is at point B is negative
Answer:
The answer is the option (a) The sign of velocity, acceleration & force on the particle when it is 3 cm away from A going towards B is positive, (c) The sign of velocity, acceleration & force on the particle when it is 4 cm away from B going towards A are negative, and (d) The sign of acceleration and force on the particle when it is at point B is negative.Explanation: (a) The velocity of the particle increases up to 0 when it is 3 cm away from A and is going from A to B, i.e., in the positive direction. Thus, the velocity is positive. Also, acceleration in SHM is towards the positive.
(b) Here, the velocity is positive and not negative since the particle is going towards B.
(c) Here, the particle is 4 cm away from B and is going towards A, i.e., the particle is going from B to A, i.e., in the negative direction. Hence, the velocity & acceleration towards the mean position O are negative.
(d) Here, force and acceleration both are negative as the particle is at B, and they both are towards O.
NCERT Exemplar Class 11 Physics Solutions Chapter 14: Very Short Answer
Exemplar Class 11 Physics Solutions Chapter 14 Oscillations: Very Short Answer gives clear and to-the-point solutions of the most vital concepts with regard to oscillatory motion and simple harmonic motion. Such solutions are used to assist the students to build on their quick recall ability, conceptual accuracy and are best to the quick revision prior to examination.
Question:14.19
Answer:
(i) At A, C, E, and G the displacement is maximum. Hence, the velocity of the oscillator will also be maximum.(ii) At B, D, F and H, the displacement of the oscillator is zero. Thus, there is no restoring force. Hence, the speed of the oscillator will be maximum.
Question:14.20
Answer:
If we displace the mass ‘m’ frpm its equilibrium position towards the right by a distance ‘x’, the spring B will be compressed by a distance x, & let kx be the force applied on the mass ‘m’ towards the left. If we apply the force kx on the mass, A will be extended by distance x towards the left, and apply force kx towards the left. Thus, the restoring force will act on the block as a net force towards the left.F = kx + kx
= 2kx
Thus, restoring force towards the left is 2kx.
Question:14.21
What are the two basic characteristics of a simple harmonic motion?
Answer:
(i)The direction of acceleration is towards the mean position, and$Acceleration \propto Displacement$ (from mean position)
(ii) The direction of force and displacement are opposite, thus, $F = -kx .$
$Restoring force \propto Displacement.$
These are the two basic characteristics of SHM.
Question:14.22
When will the motion of a simple pendulum be simple harmonic?
Answer:
Let us consider a pendulum whose,Length = l
Mass of bob = m, viz., displaced by angle = $\theta$
Now, Restoring force $= F=$-mg \sin \theta$
Also, if $\theta$ is small then
Sin $\theta$ = $\theta$
$=\frac{arc}{radius}$
$=\frac{x}{l}$
Thus,
$F=-mg\frac{x}{l}$
Or$F \propto (-x)$ …………. (since m, g & l are constants)
Thus, the simple harmonic motion for the small-angle
Question:14.21
What is the ratio of maximum acceleration to the maximum velocity of a simple harmonic oscillator?
Answer:
Let us consider $x = A \sin \omega t$ to be an SHMNow, $v = \frac{dx}{dt}$
$= A\omega \cos \omega t$
Now, for $v_{max}$,$\cos \omega t = -1$
Thus, $v_{max} = A \omega$
Now, $a = \frac{dv}{dt}$
$= -A\omega ^{2} \sin \omega t$
Now, for $a_{max}$,
$\sin \omega t = -1$
Thus, $a_{max} = A\omega ^{2}$
Now, $\frac{a_{max}}{v_{max}} = \frac{A\omega ^{2}}{A\omega }$
$=\omega$.
Question:14.24
What is the ratio between the distance travelled by the oscillator in one time period and amplitude?
Answer:
We already know that the distance travelled by an oscillator in a one-time period is equal to 4A,Where A = amplitude of oscillation.
Thus, we know that the
Required ratio
$\frac{4A}{A}$
=4:1
Question:14.25
Answer:
At time = t, P’ is the foot perpendicular to the velocity vector of particle P.
This foot shifts from P to Q, i.e., towards the negative axis, when the particle moves from P to P1.
Thus, there is –ve sign of $\theta$ motion of P’.
Question:14.26
Show that for a particle executing SHM, velocity and displacement have a phase difference of $\frac{\pi}{2}$.
Answer:
Let us consider $x = A \sin \omega t$ to be a SHM …….. (i)Now, $v = \frac{dx}{dt}$
$= A\omega \cos \omega t$
$= A\omega \sin (90^{\circ} + \omega t)$ …….. [since $\sin (90^{\circ} + \theta ) = \cos \theta$]
Thus, $v = A\omega \sin \left (\omega t + \left ( \frac{\pi}{2} \right ) \right )$
From (i), we know that,
The phase of displacement = $\omega t$
& from (ii), we know that,
The phase of velocity = $\left (\omega t + \left ( \frac{\pi}{2} \right ) \right )$
Thus, the phase difference = $\omega t + \frac{\pi}{2} - \omega t$
$= \frac{\pi}{2}.$
Question:14.27
Answer:
Let us consider that a mass is lying on a horizontal, frictionless surface, where the spring constant = k.
If we displace the mass by distance A from its mean position, then it will execute SHM.
At this stretched position, P.E. of mass = $\frac{1}{2} kA^{2}.$
Now, at maximum stretch, i.e.,
At x = A, K.E. = 0
Now, at $x = x\pm A$
P.E. = total energy
= $\frac{1}{2} kA^{2}.$
Now, let’s consider that the mass is back at its mean position, the restoring force acting on the particle will be
$P.E. = \frac{1}{2} kx^{2}$
= $\frac{1}{2} m\omega ^{2}x^{2}$
The restoring force constant of oscillator is$k = m\omega ^{2}$
When $x \pm A$
$P.E. = \frac{1}{2} m\omega ^{2}A^{2}$
& $K.E. = \frac{1}{2} mv^{2}$
$= \frac{1}{2} m\omega ^{2} (A^{2} -x^{2})$
Now, when $x = \pm A,K.E. = 0$
| x | K.E. | P.E. | T.E. |
| 0 | $\frac{1}{2} m\omega ^{2}A^{2}$ | 0 | $\frac{1}{2} m\omega ^{2}A^{2}$ |
| +A | 0 | $\frac{1}{2} m\omega ^{2}A^{2}$ | $\frac{1}{2} m\omega ^{2}A^{2}$ |
| -A | 0 | $\frac{1}{2} m\omega ^{2}A^{2}$ | $\frac{1}{2} m\omega ^{2}A^{2}$ |
E = K.E. + P.E.
= $\frac{1}{2} m\omega ^{2}A^{2}$
Thus, with displacement ‘x’, E is constant.
Question:14.28
Answer:
A second pendulum is a pendulum with a time period (T) = 2 seconds$T_{e} = 2\pi\sqrt{\frac{ l_{e}}{g_{e}}}$
Thus, $T_{e}^{2} = 4\pi ^{2}{\frac{l_{e}}{g_{e}}}$ …… (i)
Now, $T_m = 2\pi \sqrt{ \frac{l_{m}}{g_{m}}}$
Thus, $T_{m}^{2} = 4\pi ^{2} 6\frac{ lm}{g_e}$ ……. (ii) (gm=ge/6)
Now, for the second pendulum,
$\frac{T_{m}^{2}}{T_{e}^{2}}= {6g_e} \frac{l_{e}}{g_{e}l_m}$
$1 = \frac{6lm}{l_{e}}$
$1 = \frac{6l_{m}}{1m}$ ……. (since $l_{e}=1m$)
Thus, $l_{m} = \frac{1}{6} m$
NCERT Exemplar Class 11 Physics Solutions Chapter 14: Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations: Short Answer has well-organised answers to both theory and numerical questions on oscillatory motion. Such solutions assist students in consolidating theoretical knowledge, developing skills to write answers in the right format, and being ready to score well in exams.
Question:14.29
Answer:
If we pull mass M & then release it, it oscillates with the pulley up & down. Let x0 be the extension of the string when loaded with M Due to acceleration and the same amount of forces, the extension and compression of the spring from the initial position are larger and smaller, respectively. Hence, we can neglect the gravitational force here.Now, let us apply force ‘F’ to pull M downwards by a displacement x. As the string cannot be extended, its extension will be 2x.
Thus, the total extension = $(x_{0} + 2x)$
When we pull it downward by x
$T^{'}=k(x_{0} + 2x)$
When we do not pull M,
$T^{'}=kx_{0}$
F = 2T
Thus, $F=2kx_{0}$
& $F{'} = 2T{'}$
$F{'} = 2k (x_{0} + 2x)$
Now, restoring force,
Frest $= - (F^{'}-F)$
= $-[2k(x_{0} + 2x) - 2kx_{0}]$
= $-2k.2x$
Thus, $Ma = -4kx$
$a = -\frac{4k}{M }x$
Thus, $a \propto -x$
Therefore, it is a simple harmonic motion.
Now, $a = -\omega ^{2}x$
$\omega ^{2}=- \frac{a}{x}$
$= \frac{+4k}{M}$
$\omega = 2\sqrt{\frac{k}{M}}$
$T = \pi \sqrt{\frac{M}{k} }$
Question:14.3
Answer:
$y = \sin \omega t - \cos \omega t$$= \sqrt{2}\ \left [sin \omega t \frac{1}{\sqrt{2}}- \cos \omega t \frac{1}{\sqrt{2}} \right ]$
$= \sqrt{2}\ sin\left [ \omega t- \frac{\pi }{4} \right ]$
Comparing this equation with the standard SHM, we get,
$\omega = \frac{2 \pi }{T}$
Or $t = \frac{2 \pi }{\omega}$
Question:14.31
Answer:
Let us consider an oscillator, viz. adisplacement = x from its mean position,
Mass = m
$P.E. = \frac{1}{2} kx^{2}$
Force constant of oscillator, $k = m\omega ^{2}$
Thus, $P.E. = -m^{2}x^{2}$
At x = A, when K.E. = 0, P.E. will be maximum & it will be the total energy of the oscillator
$E = \frac{1}{2} m\omega ^{2}A^{2}$
At displacement ‘x’, $P.E. = \frac{1}{2}E$
Thus,
$\frac{1}{2} m\omega ^{2}x^{2}$= $\frac{1}{2}.\frac{1}{2} m\omega ^{2}A^{2}$
& $x = \pm \frac{A}{\sqrt{2}}$
Thus, when displacement = $\pm \frac{1}{\sqrt{2}}$ amplitude from mean position, P.E. will be half of the total energy.
Question:13.32
Answer:
Here, dW = F.dxSo, if W = U, dU = F.dx
Or $F = -\frac{dUx}{dx}$ ….. (since, restoring force here is opposite to displacement)
$F = -\frac{d}{dx} [U_{0} (1 - \cos \alpha x) = - \frac{d}{dx} [U_{0} + U_{0} \cos \alpha x]$
$F = -\alpha U_{0} \sin \alpha x$
Now, $\alpha x$ is small for SHM, So sin $\alpha x$ will become $\alpha x$ …… (i)
Thus, $F = -\alpha U_{0} \alpha x$
Now, since, $U_{0}$ & $\alpha$ are constants,
$F \propto -x$
Thus, the motion will be SHM.
From (ii)
$k = \alpha ^{2}U_{0}$
$m \omega ^{2}= \alpha ^{2}U_{0}$
Thus, $\omega ^{2} = \alpha ^{2} .\frac{U_{0}}{m}$
$\left ( \frac{2\pi}{T} \right )^{2} = \alpha ^{2} .\frac{U_{0}}{m}$
$T = \frac{2\pi }{\alpha }\sqrt{\frac{m}{U_{0}}}$
Thus, considering (i) the time period is valid for the small-angle $\alpha x$.
Question:14.33
Answer:
Here, the value of x is-x = 5 sin 5t.
Question:14.34
Answer:
Now, $\Theta_{1} = \Theta_{0} \sin (\omega t + \delta_{1})$& $\Theta_{2} = \Theta_{0} \sin (\omega t + \delta_{2})$
Now, for the first pendulum,
$\Theta = 2^{\circ}$
Thus, $\sin (\omega t + \delta _{1}) = 1$
& for the second pendulum,
$\Theta = -1^{\circ}$
Thus, $\sin (\omega t + \delta _{2}) = -1$
Thus, $(\omega t + \delta _{1}) = 90^{\circ}$ & $(\omega t + \delta _{2}) = 30^{\circ}$
Therefore,
$\delta _{1}+\delta _{2}=120^{\circ}$
NCERT Exemplar Class 11 Physics Solutions Chapter 14: Long Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations: Long Answer provides detailed, step-by-step explanations for in-depth questions related to oscillatory motion and simple harmonic motion. These are the solutions that enable students to gain a good conceptual clarity, logical thinking, and appropriate derivation, which is the key to good performance in descriptive exams.
Question:14.35
A person normally weighing 50 kg stands on a massless platform which oscillates up and down harmonically at a frequency of 2.0 s-1 and an amplitude 5.0 cm. A weighing machine on the platform gives the persons weight against time.
(a) Will there be any change in weight of the body, during the oscillation?
(b) If answer to part (a) is yes, what will be the maximum and minimum reading in the machine and at which position?
Answer:
Due to the normal reaction ‘N’, there will be weight in the weight machine.Let us consider the top positions of the platform, both the forces, due to the weight of the person and the oscillator act downwards.
Thus, the motion will be downwards.
Let us consider, acceleration = a
ma = mg – N …….. (i)
Now, when the platform moves upwards from its lowest position,
ma = N – mg …….. (ii)
Now, the acceleration of the oscillator is
$a = \omega ^{2}A$
From (i),
$N = mg - m \omega ^{2}A$
Where,
Amplitude = A
Angular frequency = ω &
Mass of oscillator = m
$\omega =2 \pi v$
$= 4 \pi rad /s$
A = 5cm
$= 5 \times 10^{-2}m$
m = 50 kg
$N = 50 \times 9.8 - 50 \times 4\pi \times 4\pi \times 5 \times 10^{-2}$
$= 50[ 9.8 -16\pi ^{2} \times 5\times 10^{-2}]$
$= 50 \times 1.91$
$= 95.5 N$
Thus, the minimum weight is 95.5 N.
From (ii),
N – mg = ma
Now, for upward motion from the lowest point of the oscillator,
$N = m (a + g)$
$= m [ 9.81 + \omega ^{2}A] = \omega ^{2}A$
$= 50 [9.81 + 7.89]$
$= 50 [17.7]$
$= 885 N$
Therefore, during oscillation, there is a change in the weight of the body.
Also, maximum weight = 885 N, when the platform moves upward direction from the lowest direction & minimum weight = 95.5 N, when the platform moves downward direction from the highest point.
Question:14.36
A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation?
(b) Find the frequency of oscillation?
Answer:
(a) As no deforming force acts on the spring when mass ‘m’ is supported by hand extension in the spring. Let m reach its new position at a displacement = x units from the previous one, then,P.E. of the spring or mass = gravitational P.E. lost by man
P.E. = mgx
But due to spring,
$\frac{1}{2} kx^{2}k = \omega ^{2}A$
Thus, $x = \frac{2mg}{k}$
When the extension is $x_{0}$, the spring will be at the mean position by the block
$F=kx_{0}$
F = mg
Thus, $x_{0}=\frac{mg}{k}$ ……. (ii)
Now, from (i) & (ii)
$x=2\left (\frac{mg}{k} \right )$
=$2x_{0} = 4cm$
Thus, $x_{0} = 2cm$
$x - x_{0} = 4-2 = 2cm.$
Thus, from the mean position, the amplitude of the oscillator is maximum.
(b) Now, we know that the time period (T) does not depend on the amplitude,
$T = 2\pi \sqrt{\frac{m}{k}}$
From (i)
$\frac{2mg}{k} = x$
$\frac{m}{k} = \frac{x}{2g}$
$= \frac{4 \times 10^{-2}}{2\times 9.8}$
Or, $\frac{k}{m}= \frac{2\times 9.8}{4\times 10^{-2}}$
Now, $v = \frac{1}{2}\pi \sqrt{\frac{k}{m}}$
$= \frac{1}{2}\times 3.14\sqrt{\frac{2 \times 9.8}{4 \times 10^{-2}} }$
$= \frac{10 \times 2.21}{6.28}$
$= 3.52 Hz$
Since the total extension in the spring is 4 cm when released & amplitude is 2 cm, the oscillator will not rise above 4cm, & hence, it will oscillate below the released position.
Question:14.37
A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.
$T = 2 \pi \sqrt{\frac{m}{A \rho g}}$
where m is mass of the body and ρ is density of the liquid.
Answer:
If we press a log downward into the liquid, a buoyant force acts on it, and due to inertia, it moves upwards from its mean position & comes down again due to gravity.
Thus, the restoring force on the block = Buoyant force (B.F.) – mg
Volume of liquid displaced by the block = V
When it floats,
mg = BF
Or, $mg = V\rho g$
$mg = Ax_{0} \rho g$ …….. (i)
Area of crossection = A
Height of liquid block =$x_{0}$
When pressed in water, the total height of the block in water = $(x + x_{0})$
Thus, net restoring force = $[A(x + x_{0})]\rho g -mg$
Frest $= -Ax\rho g$ …….. ( since BF is upward & x is downward)
Frest proportional to-x
Hence, the motion here is SHM.
Now, $kA\rho g$
$a = -\omega ^{2}x\omega$
$= \frac{k}{m}$
Thus, $T= 2 \pi \sqrt{\frac{m}{k}}$
Frest $= -A\rho gx$
$ma= -A\rho gx$
$a=\frac{ -A\rho gx}{m}$
$-\omega ^{2}x=\frac{ -A\rho gx}{m}$
$K = A\rho g$
$\left (\frac{2\pi 2}{T} \right )= \frac{A\rho g}{m}$
Thus,
$\frac{T}{2\pi} = \frac{m}{A\rho g}$
Thus,
$T = 2 \pi \sqrt{\frac{m}{A \rho g}}$
Question:14.38
Answer:
Let the initial height = h0 of liquid in both columns; due to the pressure difference, if the liquid in arm A is pressed by x, then the liquid in arm B will rise by x.Let us consider dx as an element of height, then,
Mass $dm = A.dx.\rho,$where, Area of a cross-section of tube = A
PE of left dm element columns = (dm) gh
& PE of dm in left columns = $dm = A.dx.\rho gx$
Thus, total PE in left column = $\int_{0}^{h}A\rho gx dx=A\rho g\left [ \frac{x^{2}}{2} \right ]^{h_{0}}_{0}$
$= A\rho g\frac{ h_{1}^{2}}{2}$
From the figure
$\sin 45^{\circ} = \frac{h_{1}}{l}$
$h_{1} = h_{2} = l \sin 45^{\circ} =\frac{1}{\sqrt{2}}$
$h_{1}^{2} = h_{2}^{2} =\frac{ l^{2}}{2}$
Thus, PE in left column = PE in right column = $A\rho g \frac{ l^{2}}{4}$
Thus, total potential energy = $A\rho g \frac{ l^{2}}{2}$
Let the element move towards the right by y units due to the pressure difference, then,
Liquid column in left arm = (l – y)
Liquid column in right arm = (l + y)
PE in left arm =$A\rho g (l - y)^{2} \sin^{2} 45^{\circ}$
PE in right arm = $A\rho g (l + y)^{2} \sin^{2} 45^{\circ}$
Total $PE = A\rho g \left ( \frac{1}{2} \right )^{2} [(l-y)^{2} + (l + y)^{2}]$
Final $PE =\frac{ A\rho g}{\sqrt{2}} [l^{2} + y^{2} - 2ly + l^{2} + y^{2} + 2ly]$
$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})$
Change PE = Final PE – Initial PE
$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})-A\rho g \frac{l^{2}}{2}$
$=\frac{ A\rho g}{2} (2l^{2}+2y^{2})$
If there is a change in velocity(v) of the liquid column,
$\Delta KE = \frac{1}{2}mv^{2}$
$m = (A.2l) \rho$
Thus, $\Delta KE = A\rho lv^{2}$
Now, change in total energy = $\frac{A\rho g}{2 }(l^{2} + 2y^{2}) + - A\rho lv^{2}$
Total change in energy,
$\Delta PE + \Delta KE = 0$
Thus,$\frac{A\rho g}{2 }(l^{2} + 2y^{2}) + - A\rho lv^{2} =0$
$\frac{A\rho}{2} [g(l^{2} + 2y^{2}) + 2lv^{2}] = 0$
$\frac{A\rho}{2}$ is not equal to 0, thus,
$[g(l^{2} + 2y^{2}) + 2lv^{2}] = 0$
Now, let us differentiate w.r.t. t,
$g \left [0 + 2 \times 2\times2 \frac{dy}{dt} \right ] + 2l.2v.\frac{dv}{dt} = 0$
$\frac{d^{2}y}{dt^{2}} + \frac{g}{l} .y= 0$
Since 4v is not equal to 0,
$\frac{d^{2}y}{dt^{2}} + \omega ^{2}y= 0$
Thus, $\omega ^{2}=\frac{g}{l}$
$\frac{2\pi }{T}=\sqrt{ \frac{g}{l}}$
Thus,
$T=2\pi\sqrt{ \frac{l}{g}}$
Question:14.39
Answer:
We know that,Acceleration due to gravity inside the Earth = g.
$g^{'} = g \left (1 - \frac{d}{R} \right )$
= $= g \left (R - \frac{d}{R} \right )$
Now, R-d = y
Thus, $g^{'} = \frac{g.y}{R}$
On both of them, force at depth ‘d’ will be –
$F = -mg^{'}$
$\frac{-mg. y}{R}$
$F \propto (-y)$
Thus, in the tunnel, the motion of the body is SHM.
We can say,
$ma = -mg^{'}$
$a =\frac{ -g}{R }y$
$\omega ^{2}y =\frac{ -g}{R}y$
Thus, $\frac{2\pi }{T} =\sqrt{ \frac{g}{R}}$
Or, $T =2\pi\sqrt{ \frac{R}{g}}$.
Question:14.40
Answer:
Let us consider the diagram at which,t = 0
$\Theta = \frac{\Theta _{0}}{2} ,\Theta = \Theta _{0} \cos \omega t$
Now, at t = t1,
$\Theta =\frac{\Theta _{0}}{2}$
Thus, $\frac{\Theta _{0}}{2} = \Theta _{0} \cos \frac{2\pi }{T}t_{1}$
…….. (Given: T = 1sec)
Thus, $\cos 2\pi t_{1} = \frac{1}{2} = \frac{cos \pi}{3}$
$2\pi t_{1} = \frac{\pi}{3}$
Or, $t_{1} = \frac{1}{6}$
$\Theta = \Theta _{0} \cos 2\pi t$
$\frac{d\Theta }{dt} = -\Theta _{0} 2\pi \sin 2\pi t$
Now, at $t = \frac{1}{6}$
$\Theta =\frac{\Theta _{0}}{2}$
$\frac{d\Theta }{dt} = -\Theta _{0} 2\pi \sin 2\pi \frac{1}{6}$
$= -\Theta _{0} 2\pi \frac{3}{\sqrt{2}}$
$= -\Theta _{0} \pi \sqrt{3}$
Now, $\omega= -\Theta _{0} \pi \sqrt{3}$
i.e., $\frac{v}{l}= -\Theta _{0} \pi \sqrt{3}$
Thus, $v= - \sqrt{3}\Theta _{0} \pi l$
By the –ve sign it is clear that the bob’s motion is towards the left.
Now,
$vx = v \cos \frac{\Theta_{0}}{2}$
=$= - \sqrt{3}\Theta _{0} \pi l \cos \frac{\Theta _{0}}{2}$
$vy = v \sin \frac{\Theta _{0}}{2} = - 3\pi \Theta _{0}l \sin \frac{\Theta _{0}}{2}$
Let H’ be the vertical distance covered by vy,
$s = ut + \frac{1}{2} gt^{2}$
$H^{'}=vyt= \frac{1}{2} gt^{2}$
$\frac{1}{2}gt^{2} + 3\pi \Theta _{0}l \sin \frac{\Theta _{0}}{2} \times t - H^{'} = 0$
Thus,$\frac{1}{2}gt^{2} + 3\pi \Theta _{0}l \frac{\Theta _{0}}{2} \times t - H^{'} = 0$ ……. (Given: $\sin \frac{\Theta _{0}}{2} =\frac{\Theta _{0}}{2}$)
Now, by the quadratic formula,
$t = \frac{-B\pm \sqrt{B^{2}-4AC}}{2A}$
$t = \frac{-\frac{\sqrt{3}\pi \theta _{0}^{2}l}{4}\pm \sqrt{(3\pi ^{2}\theta _{0}^{4}l^{2})^{2}-\frac{4.1}{2.g}H^{'}}}{\frac{2g}{2}}$
We will neglect $\theta _{0}^{4}$ and $\theta _{0}^{2}$ since $\theta _{0}$ is very small.
Thus, $t = \frac{+\sqrt{2gH^{'}}}{g } H^{'} = H + H^{"}$
Thus, $\sqrt{\frac{2H}{g}}$
H’’<<H’ as $\frac{\Theta _{0}}{2}$ is very small.
Thus, H = H’
Thus, vxt = distance covered in the horizontal direction
$X =\sqrt{ 3}\pi \theta _{0}l.\sqrt{\frac{2H}{g}}$
(Given: $\cos \frac{\theta _{0}}{2}=1$)
Thus, $X = \Theta _{0}l\pi \sqrt{\frac{6H}{g}}$
The bob was at a distance of $l\sin \frac{\theta _{0}}{2}=l \frac{\theta _{0}}{2}$ from A at the time of snapping.
Thus, the distance of Bob from A, where it meets the ground, is
$=l \frac{\Theta _{0}}{2} - X$
$=l \frac{\Theta _{0}}{2} -\Theta _{0}l\pi \sqrt{\frac{6H}{g}}$
$= l\Theta _{0}\left [ \frac{1}{2} - \pi\sqrt{ \frac{6H}{g}} \right ]$
NCERT Exemplar Class 11 Physics Solutions Chapter 14 – Oscillations: Important Concepts and Formulas
Important Concepts
1. Oscillatory Motion
Oscillatory motion is a repeated to-and-fro motion about a mean position. Examples include a swinging pendulum or a vibrating spring.
2. Simple Harmonic Motion (SHM)
SHM is a special type of oscillatory motion in which the restoring force is directly proportional to the displacement from the mean position and always directed towards it.
3. Displacement, Velocity, and Acceleration in SHM
In SHM, the variation of displacement with time is sinusoidal, the maximum velocity is at the mean position, and the maximum acceleration is at the extreme positions.
4. Time Period and Frequency
Time period refers to the time that is required to make one oscillation, and frequency refers to the number of oscillations divided by one second. They both rely on the system's nature.
5. Energy in SHM
The total mechanical energy of a system under SHM is constant, and it is always changing between kinetic and potential energy.
6. Spring–Mass System
A mass attached to a spring executes SHM when displaced from equilibrium. The motion depends on the spring constant and mass of the object.
7. Simple Pendulum
A simple pendulum consists of a point mass suspended by a light string. It does SHM at small angular displacements.
8. Simple Harmonic Motion and Uniform Circular Motion
Simple Harmonic Motion and Uniform Circular Motion are closely related types of periodic motion that are studied in physics. Where uniform circular motion can be thought of as motion uniformly at constant speed along a circular path, the simple harmonic motion may be thought of as a projection of uniform motion on a straight path. This relationship assists in understanding the sinusoidal characteristics of oscillations and is the basis of studying vibrations and waves.
9. Damped Simple Harmonic Motion
Damped SHM is a form of oscillatory motion whereby the amplitude fades with time because of resistance forces such as friction or air drag. The system is constantly losing its energy to the environment, although the system itself continues to oscillate.
9. Forced Oscillations
Forced oscillations arise when a periodic force is imposed on a system and this is constant. The system swings with the frequency of the force exerted on it and not its natural frequency because of this external force.
Important Formulas
- Displacement in SHM:
$
x=A \sin (\omega t+\phi)
$
- Velocity in SHM:
$
v=\omega \sqrt{A^2-x^2}
$
- Acceleration in SHM:
$
a=-\omega^2 x
$
- Angular frequency:
$
\omega=\frac{2 \pi}{T}
$
- Time period of spring-mass system:
$
T=2 \pi \sqrt{\frac{m}{k}}
$
- Time period of simple pendulum:
$
T=2 \pi \sqrt{\frac{l}{g}}
$
- Total energy in SHM:
$
E=\frac{1}{2} k A^2
$
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 14 Oscillations
Oscillations is a concept-rich chapter that links physics to everyday motions like swings, pendulums, and vibrations. NCERT Exemplar Class 11 Physics Solutions Chapter 14 – Oscillations help students understand these motions clearly and apply formulas confidently in numerical problems.
- The solutions discuss simple harmonic motion, restoring force, and phase in incredibly simple terms, which assist the students in understanding concepts that are otherwise difficult to understand.
- The formulas are explained with each explanation of when and how to apply them, so that there is no confusion or memorisation.
- Problems related to spring-mass systems and pendulums are solved step by step, making calculations easy to follow and reducing mistakes.
- Clear explanations aid students in visualising the manner in which displacement, velocity and acceleration vary when oscillating, which enhances comprehension.
- Exemplar questions are conceptual and exam-based, and it makes them extremely beneficial in CBSE exams as well as entrance exams such as JEE.
- The next chapter on waves is a lot easier to understand when it is known about oscillations. These are solutions that aid in establishing that good conceptual connection.
- The main concepts and methods are well explained, and the students may learn to update their knowledge and be ready for the exams in a short period of time.
Approach to Solve Exemplar Questions of Chapter 14 Oscillations
Questions based on the Oscillations chapter tend to assess the knowledge in concepts such as simple harmonic motion, time period and momentum change instead of lengthy calculations. The combination of a clear and organised approach assists the students in solving the Exemplar questions in an accurate and confident manner.
- The first is to determine whether the restoring force or acceleration is proportional to the displacement and towards the mean position. This proves the motion to be either simple harmonic.
- Determine whether the problem is related to sping mass system, a simple pendulum or SHM in general. Time period formulas are different in different systems.
- List all given quantities such as mass, spring constant, length, amplitude, or angular frequency.
- Use the right equations of displacement, velocity, acceleration or energy depending on what is asked in the question. Do not apply various formulas blindly.
- Phase constants and initial conditions in questions of time should be interpreted carefully. One little error may alter the end result.
- Work through the problem systematically and keep units consistent. The use of clear steps will minimise errors in calculation and will assist in scoring step-wise marks.
- Ensure values like time period, energy, and velocity are realistic and follow SHM conditions (e.g., maximum velocity at mean position).
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise Links provide students with a structured and organised way to study physics concepts chapter by chapter. These links make it easy to access well-explained solutions for numericals, conceptual questions, and derivations as per the latest NCERT and CBSE guidelines. They help students revise efficiently, strengthen conceptual understanding, and prepare confidently for board and competitive examinations.
NCERT Exemplar Solutions Class 11 Subject-wise
NCERT Exemplar Solutions Class 11 Subject-Wise Links offer a convenient and organised way for students to access solutions for Physics, Chemistry, and Mathematics in one place. The links enable the students to study each topic in a systematic manner with correct step-by-step directions that strictly adhere to the latest NCERT syllabus. They are suitable for rapid revision, clarity of the concepts and proper preparation for exams in all science subjects in Class 11.
NCERT Solutions for Class 11 Physics Chapter-wise
NCERT Solutions for Class 11 Physics Chapter-Wise Links help students study physics in a well-organised and systematic manner. By accessing solutions chapter by chapter, learners can easily understand concepts, numericals, and derivations as prescribed by the latest NCERT syllabus. These solutions support effective revision, strengthen fundamentals, and assist students in preparing confidently for school exams and competitive examinations.
Also, Read NCERT Solution subject-wise -
- NCERT Solutions for Class 11 Maths
- NCERT Solutions for Class 11 Chemistry
- NCERT Solutions for Class 11 Biology
Check NCERT Notes subject-wise -
- NCERT Notes Class 11 Maths
- NCERT Notes for Class 11 Physics
- NCERT Notes for Class 11 Chemistry
- NCERT Notes for Class 11 Biology
Also, Check NCERT Books and NCERT Syllabus here
Frequently Asked Questions (FAQs)
Q: Where do we find the answers to exemplar exercise questions of Class 11 Physics chapter oscillations?
A:
In this, NCERT Exemplar Class 11 Physics solutions Chapter 14 Oscillations, students can find the answers of all the exemplar questions of chapter 14.
Q: What kind of questions are asked in examinations regarding Class 11 Physics topic?
A:
The question can be of MCQ, short answer or long answer type. All these types of questions are given after each chapter, the students need to go through them once before appearing for the examination.
Q: What all topics are covered in NCERT Exemplar Class 11 Physics solutions Chapter 14?
A:
Topics such as oscillations, it’s types, SHM and its characteristic features are covered in it.
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