Simple Pendulum


A simple pendulum consists of a small, massive object (known as the "bob") suspended by a light, flexible string. When the bob is pulled to one side and released, it executes simple harmonic motion. The time period of a simple pendulum of string length $l$ is given by \begin{align} T=\frac{1}{2\pi}\sqrt{\frac{l}{g}}, \end{align} where $g$ is the acceleration due to gravity. This formula is valid for small angular displacement $\theta$.

simple pendulum

The angular displacement of a simple pendulum varies with time $t$ as \begin{align} \theta=\theta_0\sin\omega t, \end{align} where $\theta_0$ is the maximum angular displacement. The angular frequency is given by \begin{align} \omega=\frac{2\pi}{T}=\sqrt{\frac{g}{l}}. \end{align}

Problems from IIT JEE

Problem (JEE Mains 2021): $T_0$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $1/16$ times of its initial value, the modified time period is

  1. $4T_0$
  2. $T_0/4$
  3. $8\pi T_0$
  4. $T_0$

Problem (JEE Mains 2002): A child swinging on a swing in sitting position, stands up, then the time period of the swing will

  1. increase
  2. decrease
  3. remains same
  4. increases if the child is tall and decreases if the child is short

Problem (JEE Mains 2020): A simple pendulum is being used to determine the value of gravitational acceleration $g$ at a certain place. The length of the pendulum is 25.0 cm and a stop watch with 1 s resolution measures the time taken for 40 oscillations to be 50 s. The accuracy in $g$ is

  1. 2.40%
  2. 5.40%
  3. 4.40%
  4. 3.40%

Solution: The recorded length of the pendulum is $l=25.0$ cm. It also indicates the accuracy of the measuring scale, which is 0.1 cm. Thus, error in length is $\Delta l=0.1$ cm.

The time for $n=40$ oscillations is $T^\prime=nT=50$ s. The error in $T^\prime$ is $\Delta T^\prime=1$ s (stop-watch's resolution).

The acceleration due to gravity is given by \begin{align} g&=\frac{4\pi^2l}{T^2} \\ &=\frac{4\pi^2 n^2 l}{(nT)^2} \\ &=\frac{4\pi^2 n^2 l}{{T^\prime}^2}.\nonumber \end{align} Differentiate and simplify to get, \begin{align} \frac{\Delta g}{g}&=\frac{\Delta l}{l}+2\frac{\Delta T^\prime}{T^\prime} \\ &=\frac{0.1}{25}+2\;\frac{1}{50}=4.4\%.\nonumber \end{align}


  1. Simple Harmonic Motion
  2. Energy in SHM
  3. Second's Pendulum
  4. Physical Pendulum

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