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`color{blue}{star}` SIMPLE HARMONIC MOTION


`color{blue} ✍️` Consider a particle oscillating back and forth about the origin of an x-axis between the limits + A and – A as shown in Fig. 14.3.

`color{blue} ✍️` This oscillatory motion is said to be simple harmonic if the displacement x of the particle from the origin varies with time as

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

........................ (14.4)

`color{blue} ✍️` where `A, ω` and `φ` are constants.

`color{blue} ✍️` Thus, simple harmonic motion (SHM) is not any periodic motion but one in which displacement is a sinusoidal function of time. Fig. 14.4 shows what the positions of a particle executing SHM are at discrete value of time, each interval of time being T/4 where T is the period of motion.

`color{blue} ✍️`Fig. 14.5 plots the graph of `x` versus `t,` which gives the values of displacement as a continuous function of time. The quantities A, ω and φ which characterize a given SHM have standard names, as summarised in Fig. 14.6. Let us understand these quantities.

`color{blue} ✍️`The amplitutde A of SHM is the magnitude of maximum displacement of the particle. [Note, A can be taken to be positive without any loss of generality].

`color{blue} ✍️`As the cosine function of time varies from `+1` to `–1`, the displacement varies between the extremes `A` and `– A`. Two simple harmonic motions may have same `ω` and `φ` but different amplitudes A and B, as shown in Fig. 14.7 (a).

`color{blue} ✍️`While the amplitude A is fixed for a given SHM, the state of motion (position and velocity) of the particle at any time t is determined by the argument `(ωt + φ)` in the cosine function. This time-dependent quantity, `(ωt + φ)` is called the phase of the motion.

`color{blue} ✍️`The value of plase at `t = 0` is `φ` and is called the phase constant (or phase angle). If the amplitude is known, φ can be determined from the displacement at `t = 0`. Two simple harmonic motions may have the same `A` and `ω` but different phase angle `φ,` as shown in Fig. 14.7 (b).

`color{blue} ✍️`Finally, the quantity `ω` can be seen to be related to the period of motion `T`. Taking, for simplicity, `φ = 0` in Eq. (14.4), we have

`color{blue} {x(t ) = A cos ωt}`

...................... (14.5)

`color{blue} ✍️`Since the motion has a period `T, x(t)` is equal to `x (t + T)`. That is,

`color{blue} {A cos ωt = A cos ω (t + T )}`

.................... (14.6)

`color{blue} ✍️`Now the cosine function is periodic with period `2π`, i.e., it first repeats itself when the argument changes by `2π`. Therefore,

`color{purple} {ω(t + T ) = ωt + 2π}`

that is

`color{blue} {ω = 2π//T}`


`color{blue} ✍️` `ω` is called the angular frequency of SHM. Its S.I. unit is radians per second. Since the frequency of oscillations is simply `1//T, ω` is 2π times the frequency of oscillation.

`color{blue} ✍️`Two simple harmonic motions may have the same A and `φ`, but different `ω`, as seen in Fig. 14.8. In this plot the curve (b) has half the period and twice the frequency of the curve (a).

Q 3169680515

Which of the following functions of time represent (a) simple harmonic motion and (b) periodic but not simple harmonic? Give the period for each case.
(1) sin ωt – cos ωt
(2) sin2 ωt
Class 11 Chapter 14 Example 3

(a) `sin ωt – cos ωt`

`= sin ωt – sin (π/2 – ωt)`

`= 2 cos (π/4) sin (ωt – π/4)`

`= √2 sin (ωt – π/4)`

This function represents a simple harmonic motion having a period T = 2π/ω and a phase angle `(–π//4)` or (7π/4)

(b) `sin2 ωt = ½ – ½ cos 2 ωt`

The function is periodic having a period T = π/ω. It also represents a harmonic motion with the point of equilibrium occurring at ½ instead of zero.


`color{blue} ✍️`In this section we show that the projection of uniform circular motion on a diameter of the circle follows simple harmonic motion. A simple experiment (Fig. 14.9) helps us visualize this connection.

`color{blue} ✍️`Tie a ball to the end of a string and make it move in a horizontal plane about a fixed point with a constant angular speed. The ball would then perform a uniform circular motion in the horizontal plane. Observe the ball sideways or from the front, fixing your attention in the plane of motion.

`color{blue} ✍️`The ball will appear to execute to and fro motion along a horizontal line with the point of rotation as the midpoint. You could alternatively observe the shadow of the ball on a wall which is perpendicular to the plane of the circle.

`color{blue} ✍️`In this process what we are observing is the motion of the ball on a diameter of the circle normal to the direction of viewing.

`color{blue} ✍️`Fig. 14.10 describes the same situation mathematically. Suppose a particle P is moving uniformly on a circle of radius A with angular speed ω. The sense of rotation is anticlockwise. The initial position vector of the particle, i.e.,

`color{blue} ✍️`the vector `OP` at `t = 0` makes an angle of `φ` with the positive direction of x-axis. In time t, it will cover a further angle `ωt` and its position vector will make an angle of `ωt + φ` with the `+v e` xaxis.

`color{blue} ✍️`Next consider the projection of the position vector `OP` on the x-axis. This will be `OP′`. The position of `P′` on the x-axis, as the particle `P` moves on the circle, is given by

`color{purple} {x(t) = A cos (omega + phi)}`

`color{blue} ✍️`which is the defining equation of SHM. This shows that if `P` moves uniformly on a circle, its projection `P′` on a diameter of the circle executes SHM.

`color{blue} ✍️`The particle `P` and the circle on which it moves are sometimes referred to as the reference particle and the reference circle respectively.

`color{blue} ✍️`We can take projection of the motion of `P` on any diameter, say the `y`-axis. In that case, the displacement `y(t)` of `P′` on the y-axis is given by

`color{purple} { y =A sin (omega t + phi)}`

which is also an SHM of the same amplitude as that of the projection on x-axis, but differing by a phase of `π//2`.

`color{blue} ✍️`In spite of this connection between circular motion and SHM, the force acting on a particle in linear simple harmonic motion is very different from the centripetal force needed to keep a particle in uniform circular motion.
Q 3189780617

Fig. 14.10 depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case.
Class 11 Chapter 14 Example 4

(a) At t = 0, OP makes an angle of `45^o = π/4` rad with the (positive direction of ) x-axis. After time t, it covers an angle `(2pi)/T t` in the anticlockwise sense, and makes an angle of `(2pi)/T + pi/4` with the x -axis.

The projection of OP on the x-axis at time t is given by,

`color{ pink} }x (t) = A cos ( (2pi)/T t + pi/4 )}`

For T = 4 s,

`x(t) = A cos ( (2pi)/4 t + pi/4 )`

which is a SHM of amplitude A, period 4 s, and an initial phase `= pi/4`

(b) In this case at t = 0, OP makes an angle of `90^o = pi/2` with the x -axis. After a time t, it covers an angle of `(pi/2 - (2pi/T t )` with the x -axis. The projection of OP on the x-axis at time t is given by

`x(t) = B cos(pi/2 - ( 2pi)/T t )`

`=B sin ((2pi)/T t )`

For `T =30 s`

`x(t) = B sin (pi/15 t )`

Writing this as `x (t ) = B cos ( pi/15 t - pi/2 )` and comparing with Eq. (14.4). We find that this represents a SHM of amplitude B, period 30 s, and an initial phase of `- pi/2`


`color{blue} ✍️`The speed of a particle v in uniform circular motion is its angular speed ω times the radius of the circle A.

`color {blue} {v = ω A}`

................... (14.8)

`color{blue} ✍️`The direction of velocity v at a time t is along the tangent to the circle at the point where the particle is located at that instant. From the geometry of Fig. 14.11, it is clear that the velocity of the projection particle P′ at time t is

`color{blue} {v(t ) = –ωA sin (ωt + φ )}`

..................... (14.9)

`color{blue} ✍️`where the negative sign shows that v (t) has a direction opposite to the positive direction of x-axis. Eq. (14.9) gives the instantaneous velocity of a particle executing SHM, where displacement is given by Eq. (14.4). We can, of course, obtain this equation without using geometrical argument, directly by differentiating (Eq. 14.4) with respect of t:

`color {blue} { v(t) = d /(dt ) x (t)`


`color{blue} ✍️`The method of reference circle can be similarly used for obtaining instantaneous acceleration of a particle undergoing SHM. We know that the centripetal acceleration of a particle P in uniform circular motion has a magnitude `v^2//A` or `ω^2A`, and it is directed towards the centre i.e., the direction is along PO. The instantaneous acceleration of the projection particle P′ is then (See Fig. 14.12)

`color{purple} { a (t) = –ω^2A cos (ωt + φ) }`

`color{blue} {= - omega^2 x (t)}`


`color{blue} ✍️`Eq. (14.11) gives the acceleration of a particle in SHM. The same equation can again be obtained directly by differentiating velocity v(t) given by Eq. (14.9) with respect to time:

`color{blue} {a(t) = d/(dt) v (t)}`


`color{blue} ✍️`We note from Eq. (14.11) the important property that acceleration of a particle in SHM is proportional to displacement. For `x(t) > 0, a(t) < 0` and for `x(t) < 0, a(t) > 0`.

`color{blue} ✍️`Thus, whatever the value of `x` between `–A` and `A`, the acceleration a(t) is always directed towards the centre. For simplicity, let us put `φ = 0` and write the expression for `x (t), v (t) ` and `a(t) " " x(t) = A cos ωt, v(t) = – ω Asin ωt`, `a(t)=–ω2 A cos ωt` The corresponding plots are shown in Fig. 14.13.

`color{blue} ✍️`All quantitites vary sinusoidally with time; only their maxima differ and the different plots differ in phase. x varies between –A to A; v(t) varies from `–ωA` to `ωA` and a(t) from `–ω^2A` to `ω^2A`.

`color{blue} ✍️`With respect to displacement plot, velocity plot has a phase difference of π/2 and acceleration plot has a phase difference of π.

Q 3149880713

A body oscillates with SHM according to the equation (in SI units),

`x = 5 cos [2π t + π/4]`.

At t = 1.5 s, calculate the

(a) displacement, (b) speed and (c) acceleration of the body.
Class 11 Chapter 14 Example 5

The angular frequency ω of the body `= 2π s^-1` and its time period T = 1 s.

At t = 1.5 s

(a) displacement `= (5.0 m) cos [(2π s^-1) × 1.5 s + π/4]`

`= (5.0 m) cos [(3π + π/4)]`

`= –5.0 × 0.707` m

`= –3.535` m

(b) Using Eq. (14.9), the speed of the body

`= – (5.0 m)(2π s^-1) sin [(2π s^-1) × 1.5 s + π/4]`

`= – (5.0 m)(2π s^-1) sin [(3π + π/4)]`

`= 10π × 0.707 m s^-1`

`= 22 m s^-1`

(c) Using Eq.(14.10), the acceleration of the body

`= –(2π s^-1)^2 × "displacement"`

`= – (2π s^-1)^2 × (–3.535 m)`

`= 140 m s^-2`


`color{blue} ✍️`Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 14.11), the force acting on a particle of mass m in SHM is

`F (t ) = ma`

`= –mω^2 x (t )`.


`color{blue} {F (t ) = –k x (t )}`

..................... (14.13)


`color{blue} {k = mω^2}`

.............. (14.14a)


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


`color{blue} ✍️`Like acceleration, force is always directed towards the mean position - hence it is sometimes called the restoring force in SHM.

`color{blue} ✍️`To summarize the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (14.4) for displacement or by Eq. (14.13) that gives its force law. Going from Eq. (14.4) to Eq. (14.13) required us to differentiate two times. Likewise by integrating the force law Eq. (14.13) two times, we can get back Eq. (14.4).

`color{brown} {"Note"}` that the force in Eq. (14.13) is linearly proportional to x(t). A particle oscillating under such a force is, therefore, calling a linear harmonic oscillator. In the real world, the force may contain small additional terms proportional to `x^2, x^3`, etc. These then are called non-linear oscillators.
Q 3109880718

Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shown in Fig. 14.14. Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations.
Class 11 Chapter 14 Example 6

Let the mass be displaced by a small distance x to the right side of the equilibrium position, as shown in Fig. 14.15. Under this situation the spring on the left side gets elongated by a length equal to x and that on the right side gets compressed by the same length. The forces acting on the mass are then,

`color{red} {F_1 = –k x}` (force exerted by the spring on the left side, trying to pull themass towards the mean position)

`color{red} {F_2 = –k x}` (force exerted by the spring on the right side, trying to push the mass towards the mean position)

The net force, F, acting on the mass is then given by,

`color{orange} {F = –2kx}`

Hence the force acting on the mass is proportional to the displacement and is directed towards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is,

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