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?