What is the frequency of an acrobat swinging on a trapeze if the cables are approximately 2.5 m long?

The frequency of an acrobat swinging on a trapeze can be calculated using the formula for the period of a simple pendulum, which is ( T = 2\pi \sqrt{\frac{L}{g}} ), where ( L ) is the length of the cables (2.5 m) and ( g ) is the acceleration due to gravity (approximately 9.81 m/s²). The frequency ( f ) is the reciprocal of the period, given by ( f = \frac{1}{T} ). Plugging in the values, the frequency is approximately 0.63 Hz.