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NCERT Physics Notes for Class 11- Oscillations – The Simple Pendulum

NCERT Physics Notes for Class 11- Oscillations - The Simple Pendulum

NCERT Physics Notes for Class 11- Oscillations – The Simple Pendulum

NCERT Physics Notes for Class 11: A simple pendulum is just a pendulum made out of a small object attached to a light string. Ideally, it is a “point” particle attached to a massless string which is fixed to a pivot point. If the pendulum is displaced from equilibrium, it swings back and forth, and its motion is periodic. The questions we want to consider are: is the motion simple harmonic, and what is the equation for the period T of the motion?

Physics Notes for Class 11- Oscillations - The Simple Pendulum

Online Physics Notes for Class 11

To answer these questions, one starts with the equation relating forces and motion. I am going to use a different variable than the textbook. I will specify the position of the particle by the distance, along an arc, that the particle is from the equilibrium position. See the figure at the end of this section. If the length of the pendulum is l, the ratio x/l is the angle θ (in radians) that the string makes with the vertical. Most textbooks use the angle θ to specify the location of the particle, so x = lθ is the connection between the text’s θ and our x. I think this will be easier for us to make reference to the spring equation and avoid using torque.

When the pendulum is hanging vertical, x = 0, right displacement is positive x and left displacement is negative x. If an object is displaced a distance x along the arc, the component of gravity in the direction of the arc is −mgsin(θ) or −mgsin(x/l). The negative sign means that the force of gravity is a restoring force. If x > 0, sin(x/l) > 0 and the force is in the negative direction. If x < 0, sin(x/l) < 0 and the force is in the positive direction. Thus,

Fx = −mgsin( x/l )

For a force to produce simple harmonic motion, the force must be proportional to −x. The equation for the simple pendulum is NOT of this form. Here the force is proportional to −sin(x/l). Thus, the motion of a simple pendulum is NOT simple harmonic. Don’t get confused with the “sin” function. A sinusoidal restoring force does not produce perfect sinusoidal motion. Only a linear restoring force gives perfect sinusoidal motion. This is easy to demonstrate. For simple harmonic motion, the period does not depend on the amplitude. Pendula with a larger amplitude (larger xmax) have a longer period than ones with a smaller amplitude (smaller xmax).

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Approximate Sinusoidal Motion

If x/l is small, less than 0.1, then sin(x/l) ≈ x/l. This can be seen from the series expansion of sin(x/l). The expansion for sin(θ) is sin(θ) = θ−θ3/3!+θ5/5!−. . . (for θ in radians). Replacing θ = x/l gives: sin(x/l) = x/l − (x/l)3/3! + (x/l)5/5! − . . . If (x/l) < 0.1, then the second term is 0.0016 times the first, so to a pretty good approximation sin(x/l) ≈ x/l. In this case, we have

Fx ≈ − mgx/l

Thus, for small angles the restoring force is proportional to displacement, and the resulting motion is simple harmonic. The proportionality constant C is mg/l. Since the proportionality constant equals mω2, we have mω2 = mg/l, or

ω = √g/l

For small oscillations, the position x along the arc of the path is just x(t) = Asin( √q/g l t+ φ).

Since T = 2π/ω we have

T ≈ 2π √ l/g

for a simple pendulum swinging at small displacements x such that xmax/l. In terms of the angle that the string makes with the vertical the maximum angle θmax should be less than 0.1 radians.

The differential equation for x(t) can be obtained by using Fx= md2x/dt2:


m d2x/ dt2 = −mgsin( x/l )


d2x/ dt2 = −gsin( x/l)

Using the small angle approximation, we have

d2x / dt2 ≈ − g/l x

To compare with the equations in the book, just substitute x = lθ.

Also available Online NCERT Physics Notes for Class 11 on Pinterest

Watch, understand and learn Simple Pendulum, click 11th Class physics for details.

Energy considerations for Simple Harmonic Motion

To obtain an expression for the total mechanical energy that a simple harmonic oscillator has, we need an expression for the potential energy for the force acting. Let’s consider the case of the simple spring. The work done by the spring force when an object moves from the position x1 to the position x2 is:

W12 =∫x2x1 (-kx)dx

One needs to integrate since the force is not constant, but rather changes with position. The integral is:

W12 = k/2 x22 − k/2 x22

From the work-energy theorem: Wnet = ∆(K.E.). If the only force acting on the object is the force of the spring, then the net work is the work done by the spring.

In this case, we have:

k/2 x12– k/2 x22 = m/2 v22 − m/2 v12

Rearranging the terms gives:

k/2 x12 + m/2 v12 = k/2 x22 + m/2 v22

All the terms on the left are from position 1, and all the terms on the right are from position 2. Since these positions are arbitrary, we have that the expression k/2 x2 + m/2 v2is a constant of the motion. In this expression, x is the position of the object and v is the speed of the object (at the position x). As the object moves back and forth, x and v both change but the special combination k/2 x2 + m/2 v2 remains constant. We recognize the first term to be the potential energy and the second term to be the kinetic energy of the object. The sum, which is the total mechanical energy, remains constant.

Etot = k/2 x2 + m/2 v2

To determine the total mechanical energy for simple harmonic motion in general, we need to substitute for x(t) and v(t). In general x(t) = Asin(ωt + φ). The velocity is given by v(t) = dx/dt = ωAcos(ωt + φ). Substituting into the above equation, we have:

Etot = k 2 A 2 sin2 (ωt + φ) + m/2 ω2A2 cos2(ωt + φ)

However, since mω/2 = k,

we have Etot = k/2 A2 (sin2 (ωt + φ) + cos2 (ωt + φ))

Since sin2 + cos2 = 1, the equation simplifies to

Etot = k/2 A2

The total mechanical energy of a mass-spring system is k/2 times the square of the amplitude of the motion. In general, if the proportionality constant is C (Fx = −Cx), the total mechanical energy is C/2 A2.

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