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In 1851 Jean Bernard Leon Foucault demonstrated the rotation of the Earth using a pendulum 99.0 cm long. Which was set up in the paris observatory. Determine the frequency of the oscillation of the pendulum?

User Alex Link
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Final answer:

The frequency of the oscillation of a 99.0 cm long pendulum is approximately 0.282 Hz, calculated using the formula T = 2π√(l/g), where T is the period of the pendulum, l is the length of the pendulum, and g is the acceleration due to gravity.

Step-by-step explanation:

To determine the frequency of the oscillation of a pendulum that is 99.0 cm long, we can use the formula for the period of a simple pendulum:


T = 2π√(l/g)

where:

  • T is the period of the pendulum (the time it takes to complete one full oscillation)
  • l is the length of the pendulum (99.0 cm or 0.99 m in this case)
  • g is the acceleration due to gravity (approximately 9.81 m/s² on the surface of the Earth)

Plugging in the values, we get:


T = 2π√(0.99 m / 9.81 m/s²)


T ≈ 2π√(0.1009 s²)


T ≈ 2π√(0.3174 s)


T ≈ 6.2832√(0.3174 s)


T ≈ 6.2832 × 0.5634 s


T ≈ 3.54 s

Once we have the period, we can find the frequency by taking the reciprocal of the period:


f = 1/T

f ≈ 1/3.54 s

f ≈ 0.282 Hz

Therefore, the frequency of the oscillation of the pendulum is approximately 0.282 Hz.

User Mathica
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