Topic Covered



`color{blue} ✍️`Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values.

`color{blue} ✍️`In section 14.5 we have seen that the velocity of a particle executing SHM, is a periodic function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy (K) of such a particle, which is defined as

`K = 1/2 mv^2`

`= 1/2 m omega^2 A^2 sin^2 (omega t + phi)`

`color{blue} { =1/2 k A^2 sin^2 (omega t + phi)}`


`color{blue} ✍️`is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position. Note, since the sign of v is immaterial in K, the period of K is `T/2`.

`color{blue} ✍️`What is the potential energy (U) of a particle executing simple harmonic motion? In Chapter 6, we have seen that the concept of potential energy is possible only for conservative forces. The spring force `color{green} {F = –kx}` is a conservative force, with associated potential energy

`color{blue} { U = 1/2 k x^2}`


`color{blue} ✍️`Hence the potential energy of a particle executing simple harmonic motion is,

`color{purple} { U(x) = 1/2 kx^2}`

`color{blue} { =1/2kA^2 cos^2 ( omega t + phi ) }`


`color{blue} ✍️`Thus, the potential energy of a particle executing simple harmonic motion is also periodic, with period T/2, being zero at the mean position and maximum at the extreme displacements

`color{blue} ✍️`It follows from Eqs. (14.15) and (14.17) that the total energy, `E`, of the system is,

`E = U + K`

`= 1/2 kA^2 cos^2 (omega t + phi) + 1/2 kA^2 sin^2(omega t + phi )`

`color {purple} {= 1/2 kA^2[cos^2(omega t + phi) + sin^2 (omega t + phi ] }`

`color{blue} ✍️`Using the familiar trigonometric identity, the value of the expression in the brackets is unity.


`color {blue} {E =1/2 kA^2}`


`color{blue} ✍️`The total mechanical energy of a harmonic oscillator is thus independent of time as expected for motion under any conservative force. The time and displacement dependence of the potential and kinetic energies of a linear simple harmonic oscillator are shown in Fig. 14.16.

`color{blue} ✍️`Observe that both kinetic energy and potential energy in SHM are seen to be always positive in Fig. 14.16. Kinetic energy can, of course, be never negative, since it is proportional to the square of speed. Potential energy is positive by choice of the undermined constant in potential energy.

`color{blue} ✍️`Both kinetic energy and potential energy peak twice during each period of SHM. For `x = 0,` the energy is kinetic; at the extremes `x = ±A,` it is all potential energy. In the course of motion between these limits, kinetic energy increases at the expense of potential energy or vice-versa.
Q 3149080813

A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 `N m^-1`. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position.
Class 11 Chapter 14 Example 7

The block executes SHM, its angular frequency, as given by Eq. (14.14b), is

`omega = sqrt(k/m)`

`= sqrt((50Nm^-1)/(1 kg) )`

`=7.07`rad `s^-1`

Its displacement at any time t is then given by,

`x(t) = 0.1 cos (7.07t)`

Therefore, when the particle is 5 cm away from the mean position, we have

`0.05 = 0.1 cos (7.07t)`

Or `cos (7.07t) = 0.5` and hence

`sin(7.07t ) = sqrt3/2 = 0.866`

Then, the velocity of the block at x = 5 cm is

`= 0.1 × 7.07 × 0.866 m s^-1`

`= 0.61 m s^-1`

Hence the K.E. of the block,

`=1/2 mv^2`

`= 1/2 [ 1 kg xx ( 0 .6123 ms^-1)^2 ]`

= 0.19 J

The P.E. of the block,

`=1/2 kx^2`

`= ½(50 N m^-1 × 0.05 m × 0.05 m)`

= 0.0625 J

The total energy of the block at x = 5 cm,

= K.E. + P.E.

= 0.25 J

we also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system,

= ½(50 N m–1 × 0.1 m × 0.1 m )

= 0.25 J

which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of conservation of energy


`color{blue} ✍️`There are no physical examples of absolutely pure simple harmonic motion. In practice we come across systems that execute simple harmonic motion approximately under certain conditions. In the subsequent part of this section, we discuss the motion executed by some such systems.

Oscillations due to a Spring

`color{blue} ✍️`The simplest observable example of simple harmonic motion is the small oscillations of a block of mass m fixed to a spring, which in turn is fixed to a rigid wall as shown in Fig. 14.17.

`color{blue} ✍️`The block is placed on a frictionless horizontal surface. If the block is pulled on one side and is released, it then executes a to and fro motion about a mean position. Let `x = 0`, indicate the position of the centre of the block when the spring is in equilibrium.

`color{blue} ✍️`The positions marked as `–A` and `+A` indicate the maximum displacements to the left and the right of the mean position. We have already learnt that springs have special properties, which were first discovered by the English physicist Robert Hooke.

`color{blue} ✍️` He had shown that such a system when deformed, is subject to a restoring force, the magnitude of which is proportional to the deformation or the displacement and acts in opposite direction.

`color{blue} ✍️`This is known as Hooke’s law (Chapter 9). It holds good for displacements small in comparison to the length of the spring. At any time t, if the displacement of the block from its mean position is `x`, the restoring force F acting on the block is,

`color{blue} {F(x) = - kx}`


`color{blue} ✍️`The constant of proportionality, `k`, is called the spring constant, its value is governed by the elastic properties of the spring. A stiff spring has large k and a soft spring has small k. Equation (14.19) is same as the force law for SHM and therefore the system executes a simple harmonic motion. From Eq. (14.14) we have,

`color{blue} { omega = sqrt(k/m)}`


`color{blue} ✍️`and the period, `T`, of the oscillator is given by,

`color{blue} {T = 2 pi sqrt(m/k)}`


`color{blue} ✍️`Stiff springs have high value of k (spring constant). A block of small mass m attached to a stiff spring will have, according to Eq. (14.20), large oscillation frequency, as expected physically.
Q 3119191019

A 5 kg collar is attached to a spring of spring constant 500 N m–1. It slides without friction over a horizontal rod.
The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate

(a) the period of oscillation,
(b) the maximum speed and
(c) maximum acceleration of the collar.
Class 11 Chapter 14 Example 8

(a) The period of oscillation as given by Eq. (14.21) is,

`T= 2pi sqrt(m/k) = 2 pi sqrt ((5.0 kg)/(500 Nm^-1))`

`= (2π/10) s`

`= 0.63 s`

(b) The velocity of the collar executing SHM is given by,

`v(t) = –Aω sin (ωt + φ )`

The maximum speed is given by,

`v_m = Aω`

`= 0.1 xx sqrt(k/m)`

`= 0.1 xx sqrt ((500 N m^-1)/(5 kg))`

`= 1ms^-1` and it occurs at x = 0

(c) The acceleration of the collar at the displacement x (t) from the equilibrium is given by,

`a (t) = –ω^2 x(t)`

`= -k/m x (t)`

Therefore the maximum acceleration is,

`a_(max) = ω^2 A`

`= (500 Nm^-1) /(5 kg) xx 0.1 m`


and it occurs at the extremities.

The Simple Pendulum

`color{blue} ✍️`It is said that Galileo measured the periods of a swinging chandelier in a church by his pulse beats. He observed that the motion of the chandelier was periodic. The system is a kind of pendulum.

`color{blue} ✍️`You can also make your own pendulum by tying a piece of stone to a long unstretchable thread, approximately 100 cm long. Suspend your pendulum from a suitable support so that it is free to oscillate. Displace the stone to one side by a small distance and let it go.

`color{blue} ✍️`The stone executes a to and fro motion, it is periodic with a period of about two seconds. We shall show that this periodic motion is simple harmonic for small displacements from the mean position.

`color{blue} ✍️`Consider simple pendulum — a small bob of mass m tied to an inextensible mass less string of length `L`. The other end of the string is fixed to a support in the ceiling. The bob oscillates in a plane about the vertical line through the support.

`color{blue} •` Fig. 14.18(a) shows this system.

`color{blue} •` Fig. 14.18(b) is a kind of ‘free-body’ diagram of the simple pendulum showing the forces acting on the bob.

`color{blue} ✍️`Let `θ` be the angle made by the string with the vertical. When the bob is at the mean position, `θ = 0`

`color{blue} ✍️`There are only two forces acting on the bob; the tension T along the string and the vertical force due to gravity `(=mg)`.

`color{blue} ✍️`The force mg can be resolved into the component `mg cosθ` along the string and mg sinθ perpendicular to it. Since the motion of the bob is along a circle of length L and centre at the support point, the bob has a radial acceleration `(ω^2L)` and also a tangental acceleration; the latter arises since motion along the arc of the circle is not uniform.

`color{blue} ✍️`The radial acceleration is provided by the net radial force `T –mg cosθ`, while the tangential acceleration is provided by mg sinθ. It is more convenient to work with torque about the support since the radial force gives zero torque. Torque τ about the support is entirely provided by the tangental component of force

`color{blue} {τ = –L (mg sinθ ) }`


`color{blue} ✍️`This is a restoring torque that tends to reduce angular displacement — hence the negative sign. By Newton’s law of rotational motion,

`color{blue} {τ = I α}`

................. (14.23)

`color{blue} ✍️`where `I` is the moment of inertia of the system about the support and `α` is the angular acceleration. Thus,

`color{blue} {I α = –m g sin θ L}`

...................... (14.24)


`color{blue} {alpha = - (mgL) /I sin theta}`


`color{blue} ✍️`We can simplify Eq. (14.25) if we assume that the displacement `θ` is small. We know that `sin θ` can be expressed as,

`color{blue} {sin theta = theta - theta^3/(3!) + theta^5/(5!) + `


where `θ` is in radians.

`color{blue} ✍️`Now if `θ` is small, `sin θ` can be approximated by `θ` and Eq. (14.25) can then be written as,

`color{blue} {alpha = - (mgL)/I theta}`

.......................... (14.27)

`color{blue} ✍️`In Table 14.1, we have listed the angle `θ` in degrees, its equivalent in radians, and the value of the function `sin θ` . From this table it can be seen that for `θ` as large as 20 degrees, `sin θ` is nearly the same as `θ` expressed in radians.

`color{blue} ✍️`Equation (14.27) is mathematically, identical to Eq. (14.11) except that the variable is angular displacement. Hence we have proved that for small θ, the motion of the bob is simple harmonic. From Eqs. (14.27) and (14.11),

`color {purple} { omega = sqrt ((mgL)/I)}`


`color{blue} {T = 2pi sqrt (I/(mgL))}`


`color{blue} ✍️`Now since the string of the simple pendulum is mass less, the moment of inertia I is simply `mL^2`. Eq. (14.28) then gives the well-known formula for time period of a simple pendulum.

`color {blue} {T = 2pi sqrt (L/g) }`

Q 3129291111

What is the length of a simple pendulum, which ticks seconds ?
Class 11 Chapter 14 Example 9

From Eq. (14.29), the time period of a simple pendulum is given by,

`T = 2 pi sqrt (L/g)`

From this relation one gets,

`L = (gT^2)/(4 pi^2)`

The time period of a simple pendulum, which ticks seconds, is 2 s. Therefore, for g = 9.8 `m s^-2` and `T = 2 s`, L is

`= (9.8 (ms^-2) xx 4 (s^2))/(4 pi^2)`

`= 1 m`


`color{blue} ✍️`We know that the motion of a simple pendulum, swinging in air, dies out eventually. This is because the air drag and the friction at the support oppose the motion of the pendulum and dissipate its energy gradually.

`color{blue} ✍️`The pendulum is said to `color{blue} "execute damped oscillations"`. In dampled oscillations, the energy of the system is dissipated continuously; but, for small damping, the oscillations remain approximately periodic. The dissipating forces are generally the frictional forces.

`color{blue} ✍️`To understand the effect of such external forces on the motion of an oscillator, let us consider a system as shown in Fig. 14.19. Here a block of mass m connected to an elastic spring of spring constant k oscillates vertically. If the block is pushed down a little and released, its angular frequency of oscillation is `omega = sqrt (k/m)`, as seen in Eq. (14.20).

`color{blue} ✍️`However, in practice, the surrounding medium (air) will exert a damping force on the motion of the block and the mechanical energy of the block-spring system will decrease. The energy loss will appear as heat of the surrounding medium (and the block also) [Fig. 14.19].

`color{blue} {F_d = –b v}`

.................. (14.30)

`color{blue} ✍️`where the positive constant b depends on characteristics of the medium (viscosity, for example) and the size and shape of the block, etc. Eq. (14.30) is usually valid only for small velocity.

`color{blue} ✍️`When the mass m is attached to the spring and released, the spring will elongate a little and the mass will settle at some height. This position, shown by O in Fig 14.20, is the equilibrium position of the mass.

`color{blue} ✍️`If the mass is pulled down or pushed up a little, the restoring force on the block due to the spring is `color{purple} {F_S = –kx}`, where x is the displacement* of the mass from its equilibrium position. Thus, the total force acting on the mass at any time t, is `"F = –kx –bv"`.

`color{blue} ✍️`If `a(t)` is the acceleration of mass at time `t`, then by Newton’s Law of Motion applied along the direction of motion, we have

`color{blue} { m a(t) = –k x(t) – b v(t)}`

................ (14.31)

`color{blue} ✍️`Here we have dropped the vector notation because we are discussing one-dimensional motion.

`color{blue} ✍️`Using the first and second derivatives of `x (t)` for `v (t)` and `a (t)` respectively, we have

`color{blue} { m (d^2x)/(dt^2) + b (dx)/(dt) + kx = 0}`

...................... (14.32)

`color{blue} ✍️`The solution of Eq. (14.32) describes the motion of the block under the influence of a damping force which is proportional to velocity. The solution is found to be of the form

`color{blue} {x(t) = A e^(-b t//2m) cos (ω′t + φ )}`

.......................... (14.33)

`color{blue} ✍️`where a is the amplitude and `ω ′` is the angular frequency of the damped oscillator given by,

`color{blue} { omega' = sqrt(k/m - b^2/(4m^2) )}`

`color{blue} ✍️`In this function, the cosine function has a period `2π//(ω′)` but the function x(t) is not strictly periodic because of the factor `e^(-b t//2m)` which decreases continuously with time. However, if the decrease is small in one time period T, the motion represented by Eq. (14.33) is approximately periodic.

`color{blue} ✍️`The solution, Eq. (14.33), can be graphically represented as shown in Fig. 14.20. We can regard it as a cosine function whose amplitude, which is `Ae^(-b t//2m)`, gradually decreases with time.

`color{blue} ✍️`Now the mechanical energy of the undamped oscillator is `1/2 kA^2`. For a damped oscillator, the amplitude is not constant but depends on time. For small damping, we may use the same expression but regard the amplitude as `A e^(-bt/2m`.

`color{blue} {E(t) = 1/2 k A^2 e^(-bt//m)}`


`color{blue} ✍️`Equation (14.35) shows that the total energy of the system decreases exponentially with time.

`color{brown} {"Note"}` that small damping means that the dimensionless ratio `(b/sqrt(km))` is much less than 1.

`color{blue} ✍️`Of course, an expected, if we put b = 0, all equations of a damped oscillator in this section reduce to the corresponding equations of an undamped oscillator.
Q 3280101017

For the damped oscillator shown in Fig. 14.20, the mass m of the block is 200 g, `k = 90 N m^-1` and the damping constant b is 40 `g s^-1`. Calculate (a) the period of oscillation, (b) time taken for its amplitude of vibrations to drop to half of its initial value and (c) the time taken for its mechanical energy to drop to half its initial value.
Class 11 Chapter 14 Example 10

(a) We see that km `= 90 × 0.2 = 18 kg N m^-1 = kg^2 s^-2`; therefore `sqrt(km) = 4.243 kg s^-1`, and `b = 0.04 kg s^-1`. Therefore b is much less than `sqrt(km)` . Hence the time period T from Eq. (14.34) is given by

`T = 2 pi sqrt(m/k)`

`= 2pi sqrt((0.2 kg)/(90 Nm^(pm 1)))`

`= 0.3 s`

(b) Now, from Eq. (14.33), the time, `T_(1/2)`, for the amplitude to drop to half of its initial value is given by,

`T_(1//2) = ( ln(1//2))/( b//2m)`

`=( 0.693)/40 xx 2 xx 200 s`

` =6.93 s`

(c) For calculating the time, `t_(1//2)`, for its mechanical energy to drop to half its initial value we make use of Eq. (14.35). From this equation we have,

`E (t_(1//2))//E (0) = exp (–bt_(1//2)//m)`

Or ` ½ = exp (–bt_(1//2)//m)`

`ln (1/2) = –(bt_(1//2)//m)`

or `t_(1//2) = (0.693)/( 40 g s^-1) xx 200 g`

`= 3.46 s`

This is just half of the decay period for amplitude. This is not surprising, because, according to Eqs. (14.33) and (14.35), energy depends on the square of the amplitude. Notice that there is a factor of 2 in the exponents of the two exponentials.


`color{blue} ✍️`When a system (such as a simple pendulum or a block attached to a spring) is displaced from its equilibrium position and released, it oscillates with its natural frequency ω, and the oscillations are called free oscillations.

`color{blue} ✍️`All free oscillations eventually die out because of the ever present damping forces. However, an external agency can maintain these oscillations. These are called force or driven oscillations. We consider the case when the external force is itself periodic, with a frequency `ω_d` called the driven frequency.

`color{blue} ✍️`A most important fact of forced periodic oscillations is that the system oscillates not with its natural frequency `ω`, but at the frequency `ω_d` of the external agency; the free oscillations die out due to damping.

`color{blue} ✍️`A most familiar example of forced oscillation is when a child in a garden swing periodically presses his feet against the ground (or someone else periodically gives the child a push) to maintain the oscillations.

`color{blue} ✍️`Suppose an external force `F(t)` of amplitude `F_0` that varies periodically with time is applied to a damped oscillator. Such a force can be represented as

`color{blue} {F(t) = F_o cos ω_d t}`

................... (14.36)

`color{blue} ✍️`The motion of a particle under the combined action of a linear restoring force, damping force and a time dependent driving force represented by Eq. (14.36) is given by,

`color{blue} {m a(t) = –k x(t) – bv(t) + F_o cos ω_d t}`

............ (14.37a)

`color{blue} ✍️`Substituting `d^2x//dt^2` for acceleration in Eq. (14.37a) and rearranging it, we get

`color{blue} {m (d^x)/(dt^2) + b (dx)/(dt) + kx = F_o cos omega_d t}`


`color{blue} ✍️`This is the equation of an oscillator of mass m on which a periodic force of (angular) frequency `ω_d` is applied. The oscillator initially oscillates with its natural frequency `ω`.

`color{blue} ✍️`When we apply the external periodic force, the oscillations with the natural frequency die out, and then the body oscillates with the (angular) frequency of the external periodic force. Its displacement, after the natural oscillations die out, is given by

`color{blue} {x(t) = A cos (ω_dt + φ )}`


`color{blue} ✍️`where t is the time measured from the moment when we apply the periodic force.

`color{blue} ✍️`The amplitude A is a function of the forced frequency `ω_d` and the natural frequency `ω`. Analysis shows that it is given by

`color{blue} {A = F_o/(m^2 (omega^2 - omega_d^2 )^2 + omega_d^2 b^2 )^(1//2)}`



`color{blue} {tan phi = (- v_0)/(omega_d x_0)}`

.................... (14.39b)

`color{blue} ✍️`where `m` is the mass of the particle and `v_0` and `x_0` are the velocity and the displacement of the particle at time `t = 0`, which is the moment when we apply the periodic force. Equation (14.39) shows that the amplitude of the forced oscillator depends on the (angular) frequency of the driving force.

`color{blue} ✍️`We can see a different behaviour of the oscillator when `ω_d` is far from `ω` and when it is close to `ω`. We consider these two cases.

`color{brown}{(a) ul"Small Damping, Driving Frequency far from Natural Frequency :"}`
`color{blue} ✍️` In this case, `ω_d` b will be much smaller than `m(ω_2–ω_2d)`, and we can neglect that term. Then Eq. (14.39) reduces to

`color{blue} { A = F_o/( m ( omega^2 - omega_d^2 ))}`


`color{blue} ✍️`Fig. 14.21 shows the dependence of the displacement amplitude of an oscillator on the angular frequency of the driving force for different amounts of damping present in the system.

`color{blue} ✍️` It may be noted that in all the cases the amplitude is greatest when `ω_d //ω = 1`. The curves in this figure show that smaller the damping, the taller and narrower is the resonance peak.

`color{blue} ✍️`If we go on changing the driving frequency, the amplitude tends to infinity when it equals the natural frequency. But this is the ideal case of zero damping, a case which never arises in a real system as the damping is never perfectly zero.

`color{blue} ✍️`You must have experienced in a swing that when the timing of your push exactly matches with the time period of the swing, your swing gets the maximum amplitude. This amplitude is large, but not infinity, because there is always some damping in your swing. This will become clear in the (b).

`color{brown}{(b) ul"Driving Frequency Close to Natural Frequency :"}`
`color{blue} ✍️`If `ω_d` is very close to `ω , m (ω^2– 2_d^2 )` would be much less than ωd b, for any reasonable value of b, then Eq. (14.39) reduces to

`color{blue} {A= F_o/(omega_db )}`


`color{blue} ✍️`This makes it clear that the maximum possible amplitude for a given driving frequency is governed by the driving frequency and the damping, and is never infinity. The phenomenon of increase in amplitude when the driving force is close to the natural frequency of the oscillator is called resonance.

`color{blue} ✍️`In our daily life we encounter phenomena which involve resonance. Your experience with swings is a good example of resonance. You might have realised that the skill in swinging to greater heights lies in the synchronisation of the rhythm of pushing against the ground with the natural frequency of the swing.

`color{blue} ✍️`To illustrate this point further, let us consider a set of five simple pendulums of assorted lengths suspended from a common rope as shown in Fig. 14.22.

`color{blue} ✍️`The pendulums 1 and 4 have the same lengths and the others have different lengths. Now let us set pendulum 1 into motion. The energy from this pendulum gets transferred to other pendulums through the connecting rope and they start oscillating.

`color{blue} ✍️`The driving force is provided through the connecting rope. The frequency of this force is the frequency with which pendulum 1 oscillates. If we observe the response of pendulums 2, 3 and 5, they first start oscillating with their natural frequencies of oscillations and different amplitudes, but this motion is gradually damped and not sustained.

`color{blue} ✍️`Their frequencies of oscillation gradually change and ultimately they oscillate with the frequency of pendulum 1, i.e. the frequency of the driving force but with different amplitudes. They oscillate with small amplitudes. The response of pendulum 4 is in contrast to this set of pendulums.

`color{blue} ✍️` It oscillates with the same frequency as that of pendulum 1 and its amplitude gradually picks up and becomes very large. A resonance-like response is seen.

`color{blue} ✍️`This happens because in this the condition for resonance is satisfied, i.e. the natural frequency of the system coincides with that of the driving force.

`color{blue} ✍️`We have so far considered oscillating systems which have just one natural frequency. In general, a system may have several natural frequencies. You will see examples of such systems (vibrating strings, air columns, etc.) in the next chapter.

`color{blue} ✍️`Any mechanical structure, like a building, a bridge, or an aircraft may have several possible natural frequencies.

`color{blue} ✍️`An external periodic force or disturbance will set the system in forced oscillation. If, accidentally, the forced frequency `ω_d` happens to be close to one of the natural frequencies of the system, the amplitude of oscillation will shoot up( resonance), resulting in possible damage.

`color{blue} ✍️`This is why soldiers go out of step while crossing a bridge. For the same reason, an earthquake will not cause uniform damage to all building in an affected area, even if they are built with the same strength and materials.

`color{blue} ✍️`The natural frequencies of a building depend on its height, and other size parameters, and the nature of building materials. The one with its natural frequency close to the frequency of seismic wave in likely to be damaged more.