Oscillation Class 11 Notes: CBSE Physics Chapter 13

1. INTRODUCTION:

1. The periodic motion refers to the type of motion which repeats itself over and over again after regular intervals of time.

2. The oscillatory or vibratory motion refers to the type of motion in which an object moves to and fro or back and forth in a repetitive manner about a fixed point in a definite interval of time.

3. Simple harmonic motion can be considered as a specific type of oscillatory motion, in which:

1. the particle moves in a single dimension

2. the particle oscillates to and fro about a fixed mean position\[~(where~{{F}_{net}}=0),\]

3. the net force on the particle always gets directed towards the equilibrium position

4. the magnitude of the net force is always proportional to the displacement of the particle from the equilibrium position at that instant.

Mathematically, 

${{F}_{net}}=-kx$ 

where k is known as force constant

$\Rightarrow \text{ma = -- kx}$ 

$\Rightarrow a=-\frac{kx}{m}\,$

However, $a=-{{\omega }^{2}}x$ 

where, ω is known as angular frequency

\[\Rightarrow \frac{{{d}^{2}}x}{d{{t}^{2}}}=-{{\omega }^{2}}x\] 

This equation is known as the differential equation of S.H.M.

The general expression for  \[\text{x}\left( \text{t} \right)\] satisfying the above equation is:

\[x\left( t \right)=A\,\sin \,\left( \omega t+\phi  \right)\]

1.1 Some Important Terms:

1. Amplitude:

The amplitude of a particle executing S.H.M. refers to its maximum displacement on either side of the equilibrium position. The amplitude of a particle is represented by $A$.

2. Time Period:

Time period of a particle executing S.H.M. refers to the time taken to complete one cycle. It is represented by T. Mathematically,

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}\,\,\,\left( \because \omega =\sqrt{\frac{m}{k}} \right)$ 

3. Frequency:

The frequency of a particle executing S.H.M. is the same as the number of oscillations completed in one second. It is denoted by $\nu $. Mathematically,

$\nu =\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$ 

4. Phase:

The phase of a particle executing S.H.M. at any instant refers to its state with respect to its position and direction of motion at that particular instant. It is measured as argument (angle) of sine in the equation of S.H.M.

Phase$=\left( \omega t+\phi  \right)$ 

When \[t=0\], phase\[=\phi \]; the constant \[\phi \] is called the initial phase of the particle or phase constant.

1.2 Important Relations:

1. Position:

When the equilibrium position is considered at origin, the position (x coordinate) depends on time in general as \[x\text{ }\left( t \right)=\sin \text{ }\left( \omega t+\phi  \right)\]. 

• At the equilibrium position, \[\text{x = 0}\] 

• At the extremes, \[x=+a,-a\].

2. Velocity:

   
   

• At any instant t, \[v\left( t \right)=A\omega \cos \left( \omega t\text{ }+\phi  \right)\] 

• At any position x, \[v\left( x \right)=\pm \omega \sqrt{{{A}^{2}}-{{x}^{2}}}\] 

• Velocity has minimum magnitude at the extremes since the particle is at rest here. i.e., \[v=0\] at extreme position.

• On the other hand, velocity has maximum magnitude at the equilibrium position.  i.e., \[{{\left| v \right|}_{\max }}=\omega A\] at equilibrium position.

3. Acceleration:

   
   

• At any instant t, \[a\left( t \right)={{\omega }^{2}}A\text{ }\sin \left( \omega t+\phi  \right)\] 

• At any position x, \[a\left( x \right)={{\omega }^{2}}x\] 

• Acceleration is always directed towards the equilibrium position.

• The magnitude of the acceleration is minimum at the equilibrium position and maximum at extremes. ${{\left| a \right|}_{\min }}=0$ at equilibrium position ${{\left| a \right|}_{\max }}={{\omega }^{2}}A$ at extremes

4. Energy:

Kinetic Energy:

• $K=\frac{1}{2}m{{v}^{2}}\Rightarrow K=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{x}^{2}} \right)=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}{{\cos }^{2}}\left( \omega t+\phi  \right)$

• K is maximum at equilibrium position and minimum at extremes.

• ${{K}_{\max }}=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}=\frac{1}{2}k{{A}^{2}}$ at equilibrium position.

• ${{K}_{\min }}=0$ at extremes.

Potential Energy:

• If the potential energy is taken as zero at the equilibrium position, then at any position x, $U\left( x \right)=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}m{{A}^{2}}{{\omega }^{2}}{{\sin }^{2}}\left( \omega t+\phi  \right)$

• U is maximum at extremes, given by ${{U}_{\max }}=\frac{1}{2}k{{A}^{2}}$ 

• U is the minimum at the equilibrium position.

Total Energy:

• $T.E.=\frac{1}{2}k{{A}^{2}}=\frac{1}{2}m{{A}^{2}}{{\omega }^{2}}$ and is constant at all instants of time and at all positions.

 Energy Position Graph:

2. TIME PERIOD OF S.H.M:

To understand whether a motion is S.H.M. or not and to compute its time period, follow these steps:

1. Locate the equilibrium position mathematically by balancing all the forces on it.

2. Displace the particle by a displacement ‘x’ from the mean position in the probable direction of oscillation.

3. Determine the net force on it and check if it is towards the mean position.

4. Try to express net force as a proportional function of its displacement ‘x’.

• If step (c) and step (d) are proved then it is a simple harmonic motion.

5. Determine k from expression of net force \[\left( F=kx \right)\] and find the time period using $T=2\pi \sqrt{\frac{m}{k}}$. 

2.1 Oscillations of a Block Connected to a Spring:

a) Horizontal spring:

Suppose a block of mass m is placed on a smooth horizontal surface and rigidly connected to a spring of force constant K whose other end is permanently fixed.

• Mean position: when spring is at its natural length.

• Time period: $T=2\pi \sqrt{\frac{m}{k}}$ 

b) Vertical Spring:

When the spring is suspended vertically from a fixed point and carries the block at its other end as shown, the block will oscillate along the vertical line.

• Mean position: spring in elongated by $d=\frac{mg}{k}$ 

• Time period: $T=2\pi \sqrt{\frac{m}{k}}$

c) Combination of springs:

1. Springs in series:

Consider two springs of force constants \[{{K}_{1}}\] and \[{{K}_{2}}\] respectively, connected in series as shown. They are equivalent to a single spring of force constant $K$ which is given by

$\frac{1}{K}=\frac{1}{{{K}_{1}}}+\frac{1}{{{K}_{2}}}$ 

$\Rightarrow K=\frac{{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}$ 

2. Springs in parallel:

For a parallel combination as shown, the effective spring constant is \[K={{K}_{1}}+\text{ }{{K}_{2}}\]. 

2.2 Oscillation of a Cylinder Floating in a Liquid:

Suppose a cylinder of mass m and density d be floating on the surface of a liquid of density\[\rho \]. The total length of the cylinder is taken to be L.

• Mean position: cylinder is immersed up to \[\ell =\frac{Ld}{\rho }\] 

• Time period: $T=2\pi \sqrt{\frac{Ld}{\rho g}}=2\pi \sqrt{\frac{\ell }{g}}$ 

2.3 Liquid Oscillating in a U–Tube:

Consider a liquid column of mass m and density ρ in a U-tube of the area of cross-section A.

• Mean position: when the height of the liquid is the same in both limbs.

• Time period: $T=2\pi \sqrt{\frac{m}{2A\rho g}}=2\pi \sqrt{\frac{L}{2g}}$

2.4 Body Oscillation in the Tunnel Along any Chord of the Earth:

• Mean position: At the centre of the chord

• Time period: $T=2\pi \sqrt{\frac{R}{g}}=84.6$ minutes where, R is radius of earth

2.5 Angular Oscillations:

Instead of straight-line motion, when a particle or centre of mass of a body oscillates on a small arc of the circular path, then it is known as angular S.H.M. 

For angular S.H.M., torque is given by 

\[\tau =k\theta \] 

where k is a constant and $\theta $ is the angular displacement.

$\Rightarrow I\alpha =-k\theta $ 

where $I$ is the moment of inertia and $\alpha $ is the angular acceleration.

$\Rightarrow T=2\pi \sqrt{\frac{I}{k}}$, is the time period of oscillations.

2.5.1 Simple Pendulum:

• Time period: $T=2\pi \sqrt{\frac{\ell }{g}}$ 

• Time period of a pendulum in a lift:

• $T=2\pi \sqrt{\frac{\ell }{g+a}}$  (If acceleration of lift acts upwards)

• $T=2\pi \sqrt{\frac{\ell }{g-a}}$ (If acceleration of lift acts downwards)

• For a second’s pendulum:

• The time period of the second’s pendulum is 2s.

• Length of second’s pendulum on earth surface $\simeq 1m$. 

2.5.2 Physical Pendulum:

• Time period: $T=2\pi \sqrt{\frac{I}{mg\ell }}$ 

3. DAMPED AND FORCED OSCILLATIONS:

1. Damped Oscillation:

1. Damped oscillation refers to the type of vibration of a body whose amplitude keeps on decreasing with time.

2. In this type of vibration, the amplitude decreases exponentially because of damping forces like frictional force, viscous force etc.

3. Because of the decrease in amplitude, the energy of the oscillator also keeps on decreasing exponentially.

2. Forced Oscillation:

1. Forced oscillation refers to the type of vibration in which a body vibrates under the influence of an external periodic force.

2. Resonance: When the frequency of external force is the same as the natural frequency of the oscillator, then this state is called the state of resonance. This equal frequency is known as the resonant frequency.

4. WAVES:

1. Speed of Longitudinal Wave:

1. Speed of longitudinal wave in a medium mathematically represented as: $v=\sqrt{\frac{E}{\rho }}$  where E is the modulus of elasticity and \[\rho \]is the density of the medium.

2. Speed of longitudinal wave in a solid in the form of the rod is represented as $v=\sqrt{\frac{Y}{\rho }}$  where $Y$ is Young’s modulus of the solid and \[\rho \] is the density of the solid.

3. Speed of longitudinal wave in a fluid is represented as $v=\sqrt{\frac{B}{\rho }}$  where, $B$ is the bulk modulus and $\rho $ is the density of the fluid.

2. Newton’s Formula:

1. Newton considered that the propagation of the sound waves in gas is an isothermal process.  Clearly, according to him, the speed of sound in a gas is given by $v=\sqrt{\frac{P}{\rho }}$, where P is the pressure of the gas and $\rho $is the density of the gas.

2. According to Newton’s formula, the speed of sound in air at S.T.P. is \[280\text{m/s}\]. However, the experimental value of the speed of sound in air is $332m{{s}^{-1}}$. Newton could not explain this large difference during his time. In future, his formula was rectified by Laplace.

3. Laplace’s Correction:

1. Laplace considered the propagation of the sound waves in gas in an adiabatic process. Clearly, according to him, the speed of sound in a gas is given by $v=\sqrt{\frac{\gamma P}{\rho }}$, where $\gamma $ refers to the ratio of specific heats. 

2. According to Laplace’s correction, the speed of sound in air at S.T.P. is \[331.3m/s\], which agrees fairly well with the experimental values of the speed of sound in air at S.T.P.

5. WAVES TRAVELLING IN OPPOSITE DIRECTIONS:

• Consider two waves of equal amplitude and frequency propagating in opposite directions:

\[{{y}_{1}}=A\sin \left( kx\omega t \right)\] 

\[{{y}_{2}}=A\sin \left( kx\text{ }+\omega t \right)\] 

Let them be considered interfering and hence, a standing wave is produced given by,

$y={{y}_{1}}+{{y}_{2}}$ 

\[\Rightarrow y=2A\sin kx\cos \,\omega t\]

Clearly, the particle at location x is oscillating in S.H.M. with angular frequency ω and amplitude \[2A\sin kx\]. Since the amplitude depends on location (x), particles oscillate with different amplitudes.

• Nodes:

Amplitude \[=0\] 

\[\Rightarrow 2A\sin kx=0\]

$\Rightarrow x=0,\frac{\pi }{k},\frac{2\pi }{k}....$ 

$\Rightarrow x=0,\frac{\lambda }{2},\lambda ,\frac{3\lambda }{2},2\lambda ....$ 

• Antinodes: 

Amplitude is maximum.

\[\Rightarrow \sin kx=\pm 1\] 

$\Rightarrow x=\frac{\pi }{2k},\frac{3\pi }{2k}$ 

$\Rightarrow x=\frac{\lambda }{4},\frac{3\lambda }{4},\frac{5\lambda }{4}$ 

• Nodes remain at complete rest whereas antinodes oscillate with maximum amplitude (2A). The points between a node and antinode have amplitude between 0 and 2A.

• Distance between two consecutive nodes (or antinodes) \[=\frac{\lambda }{2}\]. 

• Distance between a node and the next antinode \[=\frac{\lambda }{4}\].

• Nodes and antinodes are positioned in an alternative manner.

• The figure above suggests that since nodes are at complete rest and thus, they don’t transfer energy. In a stationary wave, energy gets no transfer from one point to the other.

5.1 Vibrations in a Stretched String:

1. Fixed at Both Ends:

1. Transverse standing waves with nodes at both ends of the string are formed.

2. Clearly, length of string, $\ell =\frac{n\lambda }{2}$  if there are \[\left( \text{n + 1} \right)\] nodes and n antinodes.

3. Frequency of oscillations is given by $\nu =\frac{v}{\lambda }=\frac{nv}{2\ell }$ 

• Fundamental frequency\[\left( x=1 \right)\] or first harmonic: ${{\nu }_{0}}=\frac{v}{2L}$ 

• Second Harmonic or First Overtone: $\nu =\frac{2v}{2L}=2{{\nu }_{0}}$

• The ${{n}^{th}}$ multiple of fundamental frequency is called as ${{n}^{th}}$ harmonic or \[{{\left( n1 \right)}^{th}}\]overtone.

2. Fixed at One End:

• Transverse standing waves with node at fixed end and antinode at open end are formed.

• Clearly, length of string $\ell =\left( 2n-1 \right)\frac{\lambda }{4}$ if there are n nodes and n antinodes.

• Frequency of oscillations is given by $\nu =\frac{v}{\lambda }=\frac{\left( 2n-1 \right)v}{4\ell }$ 

• Fundamental frequency\[\left( n=1 \right)\] or first harmonic: ${{\nu }_{0}}=\frac{v}{4L}$

• First overtone or third harmonic. $\nu =\frac{3v}{4\ell }=3{{\nu }_{0}}$ 

• Only odd harmonics are possible in this case.

5.2 Vibrations in an Organ Pipe:

• Open Organ pipe (both ends open):

• The open ends of the tube form antinodes as the particles at the open end can oscillate freely.

• When there are \[\left( n+1 \right)\] antinodes in all, length of tube, \[\ell =\frac{n\lambda }{2}\]. 

• Clearly, frequency of oscillations is $\nu =\frac{nv}{2\ell }$

• Closed Organ Pipe (One end closed):

1. The open end forms an antinode and closed end forms a node.

2. When there are n nodes and n antinodes, $L=\left( 2n-1 \right)\frac{\lambda }{4}$. 

3. Clearly, frequency of oscillations is \[\nu =\frac{v}{\lambda }=\frac{\left( 2n-1 \right)v}{4L}\].

• There are only odd harmonics in a tube closed at one end.

5.3 Waves Having Different Frequencies:

Beats get formed from the superposition of two waves of slightly different frequencies propagating in the same direction. The resultant effect perceived in this case at any fixed position will consist of alternate loud and weak sounds.

Consider the net effect of two waves of frequencies ${{\nu }_{1}}$ and ${{\nu }_{2}}\,$of equal amplitude A at $x=0$.

\[{{y}_{1}}=A\,\sin \,2\pi {{\nu }_{1}}\,t\] 

${{y}_{2}}=A\,\sin \,2\pi {{\nu }_{2}}t$ 

$\Rightarrow y={{y}_{1}}+{{y}_{2}}$ 

\[\Rightarrow y=A\left( \sin 2\pi {{\nu }_{1}}\,t+\sin 2\pi {{\nu }_{2}}\,t \right)\] 

$\Rightarrow y=\left[ 2A\,\cos \,\pi \,\left( {{\nu }_{1}}\,-{{\nu }_{2}}\, \right)t \right]\,\sin \pi \left( {{\nu }_{1}}\,+{{\nu }_{2}} \right)t$ 

Clearly, the resultant wave can be denoted as a travelling wave whose frequency is

$\left( \frac{{{\nu }_{1}}\,+{{\nu }_{2}}\,}{2} \right)$ and amplitude is $2A\,\cos \,\pi \,\left( {{\nu }_{1}}\,-{{\nu }_{2}} \right)\,t$.

Since the amplitude term contains t, the amplitude varies periodically with time.

For loud sounds: 

Net amplitude $=\pm \,2A$

\[\Rightarrow \,\cos \,\pi \,\,\left( {{\nu }_{1}}\,-{{\nu }_{2}} \right)\,t=\pm \,1\] 

$\Rightarrow \pi \,\left( {{\nu }_{1}}\,-{{\nu }_{2}} \right)\,t=0,\pi ,2\pi ,3\pi ,....$ 

$\Rightarrow t=0,\frac{1}{{{\nu }_{1}}\,-{{\nu }_{2}}},\frac{2}{{{\nu }_{1}}\,-{{\nu }_{2}}},....$ 

Thus, the interval between two loud sounds represented as \[\frac{1}{{{\nu }_{1}}\,-{{\nu }_{2}}}\]. 

$\Rightarrow $ The number of loud sounds per second$={{\nu }_{1}}\,-{{\nu }_{2}}$. 

$\Rightarrow $ Beat per second $={{\nu }_{1}}\,-{{\nu }_{2}}$.

Here, it is to be noticed that ${{\nu }_{1}}\,-{{\nu }_{2}}$ should be small (0–16Hz) in order to distinguish sound variations.

Note:

• Filing a tuning fork increases its frequency of vibration whereas loading a tuning fork decreases its frequency of vibration.

6. DOPPLER EFFECT:

Doppler’s effect suggests that whenever there is a relative motion between a source of sound and a listener, the apparent frequency of sound heard by the listener is different from the actual frequency of sound emitted by the source. Mathematically, apparent frequency is given by $\nu '=\frac{\nu -{{\nu }_{L}}}{\nu -{{\nu }_{S}}}\times \nu $.

Sign Convention: 

• All velocities along the direction S to L are taken as positive and all velocities along the direction L to S are taken as negative. 

• When the motion is along some other direction, the components of velocity of source and listener along the line joining the source and listener would be considered.

Special Cases:

A. If the source is moving towards the listener but the listener is at rest, then ${{\nu }_{S}}$ is positive and ${{\nu }_{L}}=0$ (figure a). Clearly,

$\nu '=\frac{\nu }{\nu -{{\nu }_{S}}}\times \nu $ i.e., $\nu '>\nu $

B. If the source is moving away from the listener, but the listener is at rest, then ${{\nu }_{S}}$ is negative and ${{\nu }_{L}}=0$ (figure b). Clearly,

$\nu '=\frac{\nu }{\nu -\left( -{{\nu }_{S}} \right)}\times \nu =\frac{\nu }{\nu +{{\nu }_{S}}}\nu $ i.e., $\nu '<\nu $

C. If the source is at rest and listener is moving away from the source, the \[{{\nu }_{S}}=0\] and ${{\nu }_{L}}$ is positive (figure c). Clearly, 

$\nu '=\frac{\left( \nu -{{\nu }_{L}} \right)}{\nu }\nu $ i.e., $\nu '<\nu $   

D. If the source is at rest and the listener is moving towards the source, then \[{{\nu }_{S}}=0\] and ${{\nu }_{L}}$ is negative (figure d). Clearly, 

$\nu '=\frac{\nu -\left( -{{\nu }_{L}} \right)}{\nu -{{\nu }_{S}}}\nu =\frac{\nu +{{\nu }_{L}}}{\nu }\nu $ i.e., $\nu '>\nu $

E. If the source and listener are approaching each other, then \[{{\nu }_{S}}\] is positive and\[~{{\nu }_{L}}\] is negative (figure e). Clearly,

$\nu '=\frac{\nu -\left( -{{\nu }_{L}} \right)}{\nu -{{\nu }_{S}}}\nu =\frac{\nu +{{\nu }_{L}}}{\nu -{{\nu }_{S}}}\nu $ i.e., $\nu '>\nu $

F. If the source and listener are moving away from each other, then ${{\nu }_{S}}$ is negative and \[~{{\nu }_{L}}\] is positive (figure f). Clearly, 

$\nu '=\frac{\nu -{{\nu }_{L}}}{\nu -\left( -{{\nu }_{S}} \right)}\nu =\frac{\nu -{{\nu }_{L}}}{\nu +{{\nu }_{S}}}\nu $ i.e., $\nu '<\nu $

G. If the source and listener are both in motion in the same direction and with the same velocity, then ${{\nu }_{S}}={{\nu }_{L}}=\nu '$ (figure g). Clearly, 

\[\nu '=\frac{\left( \nu -\nu ' \right)}{\left( \nu -\nu ' \right)}\] i.e.,\[\nu '=\nu \] 

This suggests that there is no change in the frequency of sound perceived by the listener. Apparent wavelength perceived by the observer can be given as ${\lambda }'=\frac{\nu -{{\nu }_{S}}}{\nu }$.

Note: In case the medium is also in motion, the speed of sound with respect to the ground is also considered. i.e., $\vec{v}+{{\vec{v}}_{m}}$.

7. CHARACTERISTICS OF SOUND:

• The loudness of sound is also known as The level of intensity of sound. In decibels, the loudness of a sound of intensity I am is represented by $L=10\,{{\log }_{10}}\left( \frac{I}{{{I}_{0}}} \right)$

where, ${{I}_{0}}={{10}^{-12}}W/{{m}^{2}}$  

• The Pitch of a sound depends on its frequency. The higher the frequency of sound, the higher its pitch and the shriller it is.

Oscillations Class 11 Notes Physics - Basic Subjective Questions

Section – A (1 Mark Questions)

1. 

(a) A particle has maximum velocity in the mean 

position and zero velocity at the extreme position. Is it a sure test for S.H.M.?

(b) Is restoring force necessary in S.H.M.?

Ans. (a) No, (b)  Yes.

2. Two simple pendulums of equal lengths cross each other only at the mean position. What is their phase difference?

Ans. 180° i.e.  π radians.

3. Write the displacement equation representing the following conditions obtained in a simple harmonic motion. Amplitude = 0.01 m, frequency = 600 Hz, initial phase = π/6.

Ans. Here A = 0.01 m, $\upsilon$ = 600 Hz, φ0 = π/6

The displacement equation of simple harmonic motion is

$y=A\;sin(\omega t+\phi 0)=A\;sin(2\pi \upsilon t+\phi 0)$

or $y=(0\cdot 01m)=sin(1200\pi t+pi /6)$

4. Can a simple pendulum vibrate at the center of Earth? Why?

Ans. No. This is because the value of g at the center of Earth is zero.

5. The acceleration of a particle undergoing SHM is shown in the figure. Which of the labeled points corresponds to the particle being at –xmax ?

Ans. When particle is at $-X_{max}$ , it experiences acceleration towards mean position i.e. acceleration is positive ($+a_{max}$). So, point 1 is the correct answer.

Section – B (2 Marks Questions)

6. The motion of a particle executing SHM in one dimension is described by $x=-0\cdot 3sin\left ( t+\dfrac{\pi }{4} \right )$ , where x is in meter and t in second. Then find the frequency of oscillation in Hz.

Ans. We have, $x=-0\cdot 3sin\left ( t+\dfrac{\pi }{4} \right )$

Comparing with the general equation

$x=x_{0}sin(\omega t+\phi )$

where,$x_{0}$maximum displacement

So, $x_{0}=0\cdot 3,\omega 1,\phi =\dfrac{\pi }{4}$ .

Hence, $2\pi f=1$

$\Rightarrow f=1/2\pi$ .

7. A particle of mass 0.1 kg executes SHM under a force F = (–10x) Newton. The speed of the particle at the mean position is 6 m/s. Then find the amplitude of oscillations.

Ans. Since $f=-m\omega ^{2} x$

and at the equilibrium position, $f=-10x$

$\Rightarrow m\omega ^{2}=10$

$\Rightarrow 0\cdot 1\times \omega ^{2}=10$

$\Rightarrow \omega =10$

Also, speed at Mean position = max. Speed

$=A\omega$

$\Rightarrow 6=A\times 10$

$\Rightarrow A=0\cdot 6m$ .

8. What is the maximum acceleration of the particle doing the SHM?

$y=2sin\left [ \dfrac{\pi t}{2}+\phi  \right ]$ , where y is in cm.

Ans. $y=2sin\left [ \dfrac{\pi }{2}t+\phi  \right ]$  

Comparing with $y=A\;sin\left [ \omega t+\phi  \right ]$

$\Rightarrow $A=2$ and \omega =\dfrac{\pi }{2}$

Max. acceleration $=\omega ^{2}A=\dfrac{\pi ^{2}}{4}\times 2=\dfrac{\pi ^{2}}{2}cm/s^{2}$ .

9. A mass of 1 kg is executing SHM which is given by, x = 6.0 cos (100 t + π/4) cm. What is the maximum kinetic energy?

Ans. Here, m = 1 kg. The given equation of SHM is x = 6.0 cos (100 t + π/4) Comparing it with the equation of SHM: x = a cos (ωt + φ), 

we have, a = 6.0 cm = 6/100 m and ω = 100 rad s–1

Max. kinetic energy $=\dfrac{1}{2}m\left ( V_{max} \right )^{2}$ .

$=\dfrac{1}{2}m(a\omega )^{2}=\dfrac{1}{2}\times 1\left ( \dfrac{6}{100}\times 100 \right )^{2}=18J$ .

10. A body of mass 0.01 kg executes simple harmonic motion (S.H.M.) about x = 0 under the influence of a force shown below. Find the period of the S.H.M.

Ans. According to standard equation,

So slope of given graph

$=\dfrac{-8}{2}=-4=-m\omega ^{2}$

$F=-m\omega ^{2}x$

$m=0\cdot 01kg$

$\Rightarrow \omega ^{2}=400$

$\Rightarrow \omega =20$

and $T=\dfrac{2\pi }{\omega }=0\cdot 3sec$

Important Formula in Class 11 Physics Chapter 13 Oscillations 

1. Displacement in Simple Harmonic Motion (SHM):
$x(t) = A \cos(\omega t + \phi)$
where x(t) is the displacement, A is the amplitude, $\omega$ is the angular frequency, t is the time, and $\phi$ is the phase constant.

2. Angular Frequency:
ω=km​​
where k is the spring constant and m is the mass of the oscillating body.

3. Frequency of Oscillation:
$f= \frac{\omega}{2\pi}$
where f is the frequency and $\omega$ is the angular frequency.

4. Period of Oscillation:
$T = \frac{1}{f} = \frac{2\pi}{\omega}$
where T is the period of oscillation.

5. Maximum Speed:
$v_{\text{max}} = A \omegavma$
where $v_{\text{max}$​ is the maximum speed, A is the amplitude, and $\omega$ is the angular frequency.

6. Maximum Acceleration:
$a_{\text{max}} = A \omega^2$
where $a_{\text{max}}$​ is the maximum acceleration, A is the amplitude, and $\omega$ is the angular frequency.

Important Topics of Class 11 Physics Chapter 13 Oscillations 

Importance of Class 11 Physics Oscillations Notes

• Oscillations form a fundamental concept in Physics, crucial for understanding more advanced topics like waves and quantum mechanics. 

• These notes provide a clear foundation in simple harmonic motion and other types of oscillatory systems.

• Oscillations Class 11 Notes PDF Download simplify complex concepts related to oscillations, such as amplitude, frequency, and phase, making it easier for students to grasp these important ideas and their applications.

• By including essential formulas and equations, the notes help students learn how to calculate and analyse oscillatory motion, which is critical for solving problems and understanding practical scenarios.

• With practice questions and detailed solutions, Oscillations Class 11 Notes PDF enhance students' problem-solving skills, allowing them to apply theoretical concepts to practical problems effectively.

• Diagrams and visual representations in the notes aid in visualising oscillatory motion and related phenomena, making it easier to understand and remember the concepts.

• Class 11 Physics Oscillations Notes are prepared to align with the CBSE syllabus, helping students prepare efficiently for exams by focusing on key topics and providing a structured review of important material.

Tips for Learning the Class 11 Chapter 13 Physics Oscillations

• Start by grasping the fundamental concepts of oscillations, such as what makes a motion oscillatory and the meaning of terms like amplitude and frequency.

• Draw diagrams of oscillatory motion and use graphs to visualise how displacement, velocity, and acceleration change over time.

• Familiarise yourself with key formulas related to oscillations and practice using them in different problems to reinforce your understanding.

• Work through various practice problems to apply the concepts and formulas. This will help you get comfortable with solving oscillation-related questions.

• Look at solved examples to see how the concepts and formulas are used in real problems. Pay attention to the steps and methods used.

• Relate the concepts of oscillations to real-world examples, such as pendulums or springs, to better understand their practical applications.

• Make brief notes summarising the main ideas and formulas from the chapter. Reviewing these summaries regularly can help reinforce your learning.

• If you're unsure about any concept or formula, don’t hesitate to ask for help from teachers or classmates. Clarifying doubts early can prevent confusion later.

Conclusion 

In Class 11 Physics, the chapter on Oscillations explores the behaviour of repetitive motions, such as those seen in pendulums and springs. Grasping fundamental concepts like amplitude, frequency, and period is essential for understanding how these systems oscillate. Mastering key formulas for displacement, velocity, and acceleration is crucial for solving problems and predicting the motion of oscillatory systems. Applying these concepts through practice problems and relating them to real-life examples can significantly enhance your understanding. Regular review and summarization of the chapter’s main points help in retaining information and preparing effectively for exams. Additionally, seeking clarification on any doubts from teachers or peers can further solidify your grasp of the topic and improve problem-solving skills.

Related Study Materials for Class 12 Physics Chapter 13

Chapter-wise Links for Physics Notes For Class 11 PDF FREE Download

Related Study Materials Links for Class 11 Physics

Along with this, students can also download additional study materials provided by Vedantu for Physics Class 11–