4. Find the length of a seconds' pendulum at a place
where g = 10 m s-2 (Take a = 3-14).
Answers
Answer:
Your answer in above image-.-
Explanation:
The larger point is that the mathematics of pendulums are in general quite complicated. The differential equation which represents the motion of a simple pendulum is not, actually, easily solved. There is no solution that can be written in terms of elementary functions.
Yet, simplifying assumptions can be made. One reads the idea, that the motion of a simple pendulum is like simple harmonic motion in that the equation for the angular displacement is the same form as the motion of a mass on a spring. And this is based on adding a restriction to the size of the oscillation's amplitude. So okay, we can fudge, and this yields the equation for a harmonic oscillator. This involves us in some sort of error due to the approximation, but let’s proceed anyways..
We are proceeding anyways, and okay, trust me then, there are equations, that show how to find the frequency and period of the motion. And knowing the length of the pendulum, you can determine its frequency. But you want to determine the necessary length, from a specific frequency. And you can do this too. The ‘period’, which is of the motion for a pendulum, or, how long it takes to swing back-and-forth, measured in seconds, is given as ‘1.5 seconds’. Fine, and we plug this into a whole deal where:
T is the period in seconds (s)
π is the Greek letter pi and is approximately 3.14
√ is the square root of what is included in the parentheses
L is the length of the rod or wire in meters or feet
g is the acceleration due to gravity (9.8 m/s² or 32 ft/s² on Earth)
And there is this equation, which by the way, you can solve for L:
f = [√(g/L)]/2π
Frequency f is the reciprocal of the period T.
2πf = √(g/L)
Square both sides of the equation:
4π^2f^2 = g/L
Solve for L:
L = g/(4π^2f^2)
Again, frequency f is the reciprocal of the period T, so okay, the length equation is:
L = gT^2/4π^2
And that’s .559 or so meters, I think.
