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------------------------------------------------ A trapeze artist swings in simple harmonic motion with a period of 3.8 s. Calculate the length of the cables supporting the trapeze.

Calculate the period and frequency of a 3.500 m long pendulum at the following locations:

a. The North Pole, where [tex]a_g = 9.832 \, \text{m/s}^2[/tex]
b. Chicago, where [tex]a_g = 9.803 \, \text{m/s}^2[/tex]
c. Jakarta, Indonesia, where [tex]a_g = 9.782 \, \text{m/s}^2[/tex]

Answer :

Final answer:

The length of the cables supporting the trapeze is approximately 2.75 meters. In locations with different values of acceleration due to gravity, the period and frequency of a pendulum can be calculated using the formula T = 2π√(L/g).

Explanation:

The length of the pendulum can be determined using the simple harmonic motion formula for the period, T. The formula is given by:

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. Rearranging the formula, we can solve for L as:

L = (T/2π)^2 × g

Substituting the given period of 3.8 s, we find:

L = (3.8 s / (2π))^2 × 9.80 m/s2 = 2.75 m

Therefore, the length of the cables supporting the trapeze is approximately 2.75 meters.

For locations with different values of acceleration due to gravity, the formula for the period of a simple pendulum remains the same, but the value of g changes. Using the given lengths of 3.500 m and the values of acceleration due to gravity at each location, we can calculate the period and frequency of the pendulum.

a. At the North Pole where g = 9.832 m/s2:

T = 2π√(L/g) = 2π√(3.500 m / 9.832 m/s2) = 6.068 s

Frequency = 1/T = 1/6.068 s = 0.165 Hz

b. In Chicago where g = 9.803 m/s2:

T = 2π√(L/g) = 2π√(3.500 m / 9.803 m/s2) = 6.057 s

Frequency = 1/T = 1/6.057 s = 0.165 Hz

c. In Jakarta, Indonesia where g = 9.782 m/s2:

T = 2π√(L/g) = 2π√(3.500 m / 9.782 m/s2) = 6.044 s

Frequency = 1/T = 1/6.044 s = 0.165 Hz

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