Find the length of a pendulum that oscillates with a frequency of 0.25 Hz
0.04
To find the length of a pendulum that oscillates with a frequency of 0.25 Hz, we can use the formula for the period of a pendulum.
The formula for the period of a pendulum is:
T = 2π√(L/g)
Where:
T = period (time for one complete oscillation)
L = length of the pendulum
g = acceleration due to gravity (approximately 9.8 m/s^2 on Earth)
Since we are given the frequency (f), which is the reciprocal of the period (f = 1/T), we can rewrite the period formula as:
T = 1/f
Now, we can substitute this into the period formula:
1/f = 2π√(L/g)
Solving for L, we get:
L = (g/4π^2) * (1/f)^2
Substituting the values:
g = 9.8 m/s^2
f = 0.25 Hz
L = (9.8 / (4π^2)) * (1/0.25)^2
Calculating this expression gives:
L ≈ 9.8 / (4 * 3.14^2) * 16
Simplifying further:
L ≈ 9.8 * 16 / (4 * 9.87)
L ≈ 156.8 / 39.48
L ≈ 3.973 meters
Therefore, the length of a pendulum that oscillates with a frequency of 0.25 Hz is approximately 3.973 meters.
To find the length of a pendulum that oscillates with a frequency of 0.25 Hz, we can use the formula for the period of a pendulum:
T = 1/frequency
where T is the period and frequency is measured in Hz (1/seconds).
In this case, we are given the frequency as 0.25 Hz. Plugging this into the formula, we can calculate the period:
T = 1/0.25
T = 4 seconds
The period gives us the time it takes for one complete cycle of the pendulum. To find the length of the pendulum, we can use the formula for the period of a pendulum:
T = 2π√(L/g)
where L is the length of the pendulum and g is the acceleration due to gravity (approximately 9.8 m/s²).
Rearranging the formula to solve for L, we get:
L = (T^2 * g)/(4π^2)
Plugging in the values we have:
L = (4^2 * 9.8)/(4π^2)
L = (16 * 9.8)/(4π^2)
L ≈ 1.006 meters
Therefore, the length of the pendulum that oscillates with a frequency of 0.25 Hz is approximately 1.006 meters.
Period=2PI sqrt(L/g)
1/frequency=1/2PI * sqrt(g/L)