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on an alien planet, a simple pendulum of length 0.500 m has oscillation frequency 0.609 hz . find the period of the pendulum.

User Sosborn
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1. The period of the pendulum on the alien planet is approximately 1.410 seconds.

2. The acceleration due to gravity on the planet's surface is approximately 7.938 m/s².

3. The required length of the pendulum for matching the frequency of the spring-mass system is approximately 0.632 meters.

1. Period of the pendulum:

Frequency (f) = 0.709 Hz

Period (T) = 1/f

T = 1/0.709 Hz ≈ 1.410 s

Therefore, the period of the pendulum on the alien planet is approximately 1.410 seconds.

2. Acceleration due to gravity:

Length of pendulum (L) = 0.400 m

Period (T) = 1.410 s

Gravity (g) = 4π² * L / T²

g = 4π² * 0.400 m / (1.410 s)² ≈ 7.938 m/s²

Therefore, the acceleration due to gravity on the planet's surface is approximately 7.938 m/s².

3. Length of pendulum for matching frequency:

Desired frequency (f_desired) = 0.709 Hz

Mass of glider (m) = 0.450 kg

Spring constant (k) = 9.75 N/m

Gravity (g) = 7.938 m/s²

The frequency of a spring-mass system is given by:

f = (1/2π) * √(k/m)

To match the desired frequency with the pendulum, we need to set the two equations equal:

0.709 Hz = (1/2π) * √(k/m)

Solving for the length of the pendulum (L):

L = g / (4π² * f_desired²)

L = 7.938 m/s² / (4π² * 0.709 Hz²) ≈ 0.632 m

Therefore, the required length of the pendulum for matching the frequency of the spring-mass system is approximately 0.632 meters.





The probable question can be: On an alien planet, a simple pendulum of length 0.400 m has oscillation frequency 0.709 Hz. Find the period of the pendulum. Find the acceleration due to gravity on this planet's surface. What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass 0.450 kg attached to a spring of force constant 9.75 N/m?

User Manogna Mujje
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8.2k points