Near the surface of Earth a certain spring oscillates up and down with the same frequency that a pendulum swings back and forth. The spring holds a 7.00 kg mass and has a spring constant of 88.8 N/m. What is the length of the pendulum?
The frequency of oscillation for a spring-mass system can be calculated using the equation:
f = 1 / (2π) * √(k / m),
where f is the frequency, k is the spring constant, and m is the mass.
In this case, the frequency of the spring is the same as the frequency of the pendulum, so we can set these two equations equal to each other:
f_pendulum = f_spring
This means that the period of oscillation for the pendulum, T_pendulum, is the reciprocal of the frequency:
T_pendulum = 1 / f_pendulum
On the other hand, the period of oscillation for a pendulum can be calculated using the equation:
T_pendulum = 2π * √(L / g),
where L is the length of the pendulum and g is the acceleration due to gravity.
Setting these two equations equal to each other, we can solve for the length of the pendulum:
2π * √(L / g) = 1 / f_pendulum
Multiplying both sides by f_pendulum:
2π * f_pendulum * √(L / g) = 1
Squaring both sides to eliminate the square root:
(2π * f_pendulum * √(L / g))^2 = 1^2
4π^2 * f_pendulum^2 * (L / g) = 1
Rearranging the equation to solve for L:
L = (1 / (4π^2 * f_pendulum^2)) * g
Now, we can plug in the given values:
k = 88.8 N/m
m = 7.00 kg
Calculating the spring frequency using the equation mentioned at the beginning:
f_spring = 1 / (2π) * √(k / m)
= 1 / (2π) * √(88.8 N/m / 7.00 kg)
Plugging in the values and evaluating the expression:
f_spring ≈ 1.092 Hz
Now, calculating the length of the pendulum using the equation derived above:
L = (1 / (4π^2 * f_pendulum^2)) * g
= (1 / (4π^2 * f_spring^2)) * g
Plugging in the values for f_pendulum and g, where g is approximately 9.8 m/s^2:
L ≈ (1 / (4π^2 * (1.092 Hz)^2)) * 9.8 m/s^2
Evaluating the expression:
L ≈ 0.807 m
Therefore, the length of the pendulum is approximately 0.807 meters.