A pendulum oscillates at a frequency of 2 cycles per second. At t = 0 it is released starting at an angle of 20°C. Ignoring friction what will be the position (angle) of the pendulum at
a. t = 0.25 sec
b. t = 1.6 sec
c. t = 5.0 sec
To determine the position (angle) of the pendulum at different times, we need to understand the behavior of the pendulum and its oscillation pattern.
The period (T) of a pendulum is the time it takes to complete one full oscillation. In this case, the frequency is given as 2 cycles per second, so the period can be calculated as 1/frequency, which is 1/2 = 0.5 seconds.
For a simple pendulum, the equation of motion for the angle (θ) as a function of time (t) can be described using the formula:
θ(t) = θ_max * sin(2πft + φ)
where:
- θ(t) is the angle of the pendulum at time t
- θ_max is the maximum angle (initial angle) at t = 0
- f is the frequency (cycles per second)
- φ is the phase constant
a. t = 0.25 sec:
To find the position at t = 0.25 sec, we can substitute the values into the equation as follows:
θ(0.25) = θ_max * sin(2πft + φ)
= 20 * sin(2π * 2 * 0.25 + φ)
Since the phase constant (φ) is not given, we can assume it to be 0. Plugging in the values, we get:
θ(0.25) = 20 * sin(2π * 2 * 0.25 + 0)
= 20 * sin(2π * 0.5)
= 20 * sin(π)
= 20 * 0
= 0
Therefore, the position (angle) of the pendulum at t = 0.25 sec is 0 degrees.
b. t = 1.6 sec:
Using the same formula:
θ(1.6) = θ_max * sin(2πft + φ)
= 20 * sin(2π * 2 * 1.6 + 0)
Plugging in the values, we get:
θ(1.6) = 20 * sin(2π * 3.2)
= 20 * sin(6.4π)
≈ 20 * sin(0.2π)
≈ 20 * sin(36°)
≈ 20 * 0.588
≈ 11.76
Therefore, the position (angle) of the pendulum at t = 1.6 sec is approximately 11.76 degrees.
c. t = 5.0 sec:
Similarly,
θ(5.0) = θ_max * sin(2πft + φ)
= 20 * sin(2π * 2 * 5 + 0)
Plugging in the values, we get:
θ(5.0) = 20 * sin(2π * 10)
= 20 * sin(20π)
≈ 20 * sin(6.28π)
≈ 20 * sin(0π)
≈ 20 * 0
= 0
Therefore, the position (angle) of the pendulum at t = 5.0 sec is 0 degrees.