Simple harmonic motion I
Motion of a simple Pendulum

The period, T of a simple pendulum is given as: $$ T = 2 \pi {\sqrt{\frac{l}{g}}} $$ Where T = period in seconds (s)
l = length of pendulum in metres (m)
g = acceleration due to gravity in m/s²

For two pendulums with different lengths l1 and l2 and whose periods of oscillation T1 and T2 are different, the relationship between their period and length is given as: $$\frac{T_1}{T_2} = {\sqrt{\frac{l_1}{l_2}}} $$ $$ \frac{T_1^2}{T_2^2} = \frac{l_1}{l_2} $$

Calculations

Example 1: Calculate the length of a simple pendulum that oscillates at a frequency of 0.4Hz [g = 10m/s², π = 3.142 ] (NECO)

Solution

$$ f = 0.4Hz $$ $$ T = 2π{\sqrt{\frac{l}{g}}} $$ $$ \text{But, } T = \frac{1}{f} $$ $$ T = \frac{1}{0.4} = 2.5s $$ $$ \text{Making l subject of formula} $$ $$ T² = (2π)² × \frac{l}{g} $$ $$ l = \frac{gT²}{4π²} $$ $$ l = \frac{10 × 2.5²}{4 × 3.142²} $$ $$ l = 1.58m $$

Example 2: A simple pendulum has a period of 17.0s. When the length is shortened by 1.5m, it's period is 8.5s. Calculate the original length of the pendulum.

Solution

$$ \text{Let the original length, l} = l_1 $$ $$ \text{period at }{T_1} = 17s $$ $$ \text{new length }{l_2} = l_1 - 1.5 $$ $$ \text{period at new length }{T_2} = 8.5s $$ $$ \frac{T_1^2}{T_2^2} = \frac{l_1}{l_2} $$ $$ \frac{17²}{8.5²} = \frac{l_1}{l_1-1.5} $$ $$ 4 = \frac{l_1}{l_1-1.5} $$ $$ \text{cross multiply} $$ $$ l_1 = 4({l_1-1.5}) $$ $$ l_1 = 4l_1 - 6 $$ $$ 4l_1 - l_1 = 6 $$ $$ 3l_1 = 6 $$ $$ l_1 = \frac{6}{3} = 2m $$

Example 3: If in a simple pendulum experiment the length of the inextensible string is increased by a factor of four, its period is increased by a factor of? (JAMB)

Solution

$$ T = 2 \pi {\sqrt{\frac{l}{g}}} $$ $$ T \alpha \sqrt{l} $$ $$ \text{Keeping all factors constant} $$ $$ T = \sqrt{l} $$ $$ l = 4 $$ $$ T = \sqrt{4} $$ $$ T = 2 $$ $$ \text{The period is increased by a factor of 2} $$

Example 4: A simple pendulum, 0.6m long, has a period of 1.5s. What is the period of a simple pendulum 0.4m long in the same location?

Solution

$$ \frac{T_1^2}{T_2^2} = {\frac{l_1}{l_2}} $$ $$ T_2^2 = \frac{T_1^2 × l_2}{l_1} $$ $$ T_2^2 = 1.5 ^2 × \frac{0.4}{0.6}$$ $$ T_2 = 1.5 × {\sqrt{\frac{0.4}{0.6}}} $$ $$ T_2 = 1.5{\sqrt{\frac{2}{3}}}s $$

Period in a simple pendulum Calculator







Formulas:
  1. For 1: \( T = 2\pi \sqrt{\frac{l}{g}} \)

  2. For 2: \( T = \frac{1}{f} \)

  3. For 3: \( f = \frac{1}{T} \)

  4. For 4: \( T_2 = T_1 \times {\sqrt{\frac{l_2}{l_1}}} \)

  5. For 5: \( l_2 = \frac{T_2^2 \times l_1}{T_1^2} \)

  6. For 6: \( l = \frac{gT²}{{4 \pi}²} \)
Select the formula to use the calculator. Input the values for the parameters to solve using the calculator. Use the information below as a guide.
Simple pendulum Experiment