Question

How long must a simple pendulum be if it is to make exactly one swing per five seconds

Answer 1

Answer:

  L = 6.2 m

Explanation:

The angular velocity of a simple pendulum is

         w = √ g / L

Where w is the angular velocity, g is the acceleration of gravity, which we assume 9.8 m / s2 and L is the length of the pendulum

Angular velocity, frequency and period are related

        w = 2π f = 2 π / T

       2π / T = √ g / L

      L = g T² / 4π²

Let's calculate

      L = 9.8 5²2 / 4π²

      L = 6.2 m

Answer 2

Answer:

The simple pendulum must be 6.2 m to make one swing per five seconds.

Explanation:

The angular frequency of the simple pendulum is

$\omega = \sqrt{\frac{g}{L}}$

The period of the pendulum is then

$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L}{g}}$

The definition of the period is the time it takes to complete one cycle. So, it order the pendulum to complete one cycle in five seconds, then T = 5.

$T = 2\pi \sqrt{\frac{L}{g}}\\\frac{L}{g} = (\frac{T}{2\pi})^2\\L = \frac{T^2g}{4\pi^2} = \frac{5^2(9.8)}{4\pi^2} = 6.2~m$