What is Simple Pendulum?
A point mass attached to a light inextensible string and suspended from fixed support is called a simple pendulum. The vertical line passing through the fixed support is the mean position of a simple pendulum.
The vertical distance between the point of suspension and the centre of mass of the suspended body (when it is in mean position) is called the length of the simple pendulum denoted by L.
Table of Content:
- Important Terms
- Time Period of Simple Pendulum
- Energy of Pendulum
- Physical Pendulum
A simple pendulum is a mechanical arrangement that demonstrates periodic motion. The simple pendulum comprises of a small bob of mass ‘m’ suspended by a thin string secured to a platform at its upper end of length L.
- The oscillatory motion of a simple pendulum: Oscillatory motion is defined as the to and fro motion of the pendulum in a periodic fashion and the centre point of oscillation known as equilibrium position.
- The time period of a simple pendulum: It is defined as the time taken by the pendulum to finish one full oscillation and is denoted by “T”.
- The amplitude of simple pendulum: It is defined as the distance travelled by the pendulum from the equilibrium position to one side.
- Length of a simple pendulum: It is defined as the distance between the point of suspension to the centre of the bob and is denoted by “l”.
⇒ Also Read:
Time Period of Simple Pendulum
A point mass M suspended from the end of a light inextensible string whose upper end is fixed to a rigid support. The mass displaced from its mean position.
- There is negligible friction from the air and the system
- The arm of the pendulum does not bend or compress and is massless
- The pendulum swings in a perfect plane
- Gravity remains constant
Time Period of Simple Pendulum Derivation
Using the equation of motion, T – mg cosθ = mv2L
The torque tending to bring the mass to its equilibrium position,
τ = mgL × sinθ = mgsinθ × L = I × α
For small angles of oscillations sin ≈ θ,
Therefore, Iα = -mgLθ
α = -(mgLθ)/I
– ω02 θ = -(mgLθ)/I
ω02 = (mgL)/I
ω20 = √(mgL/I)
Using I = ML2, [where I denote the moment of inertia of bob]
we get, ω0 = √(g/L)
Therefore, the time period of a simple pendulum is given by,
T = 2π/ω0 = 2π × √(L/g)
Energy of Simple Pendulum
From the above figure, the potential energy (PE) of the bob at B with respect to A is,
m×g×h = mg × (L – L cosθ)h = L – L cosθ
= mgL × (1 – cosθ)cosθ = 1 – 2 sin2θ/2
= mgL [1 – (1 – 2 sin2 θ/2)]
For small angles sin (θ/2) ≈ θ/2
= mg × L2(θ/2)2= 1/2 (mgLθ2)
Potential Energy of a simple pendulum is1/2 (mgLθ2).
Kinetic Energy of the Bob:
1/2 (Fω2) = 1/2 (mL2) (dθ/dt)2 [ω = dθ/dt]
=1/2 (ML2) [θ20 ω02 cos2 (ω0t)]
=1/2 (ML2) ω02 [θ02 (1 – sin2 ω0t}]
=1/2 (ML2) × (g/L) × [θ02 – θ2]
∴ Kinetic Energy of a simple pendulum is [1/2 × mgLθ02] – 1/2 mgLθ2θ0 – Amplitude
Mechanical Energy of the Bob:
E = KE + PE= 1/2 mg Lθ02 =constant
The energy of the simple pendulum is conserved and is equal in magnitude to the potential energy at the maximum amplitude.
- If the temperature of a system changes then the time period of the simple pendulum changes due to change in length of the pendulum.
- A simple pendulum is placed in a non-inertial frame of reference (accelerated lift, horizontally accelerated vehicle, vehicle moving along an inclined plane).
The mean position of the pendulum may change. In these cases, g is replaced by “g effective”. For determining the time period (T).
- A lift moving upwards with acceleration ‘a’, then, T = 2π × (L/geff) = 2π √[L/(g + a)]
- If the lift is moving downward with acceleration ‘a’, then T = 2π × (L/geff), geff = √(g2 + a2) or, T = 2π × √[L/(g2 + a2)1/2]
- For simple pendulum of length ‘L’ comparable to the radius of the earth ‘R’, then the time period T = 2π [1/(g/L + g/R)]
- For infinitely long pendulum L > > R near the earth surface, T = 2π × √(R/g)
A simple pendulum is an idealized model. It is not achievable in reality. But the physical pendulum is a real pendulum in which a body of finite shape oscillates. From its frequency of oscillation, we can calculate the moment of inertia of the body about the axis of rotation.
Consider a body of irregular shape and mass (m) is free to oscillate in a vertical plane about a horizontal axis passing through a point, weight mg acts downward at the centre of gravity (G).
If the body displaced through a small angle (θ) and released from this position, a torque is exerted by the weight of the body to restore to its equilibrium.
τ = -mg × (d sinθ)
τ = I α
τ α = – mgdsinθ
I = d2θ/dt2 = – mgdsinθ
Where I = moment of inertia of a body about the axis of rotation.
d2θ/dt2 = (mgd/I) θ [Since, sinθ ≈ θ]
ω0 = √[mgd/I].
Time Period of Physical Pendulum
T = 2π/ω0 = 2π × √[I/mgd]
For ‘I’, applying parallel axis theorem,
I = Icm + md2
Therefore, the time period of a physical pendulum is given by,
T = 2π × √[(Icm + md2)/mgd]