證明單擺週期在擺角西搭<5度時為T=2拍*根號L/g

用戶37276

2013-01-27 12:12 am

證明單擺週期在擺角西搭<5度時為T=2拍*根號l/g

更新1:

x=lsin西搭當西搭很小時(sin趨近於西搭) x=l西搭 mgsin西搭=mg西搭 =mg*x/ =kx k=mg/l T=2拍*根號m/k =2拍*根號m/mg/l =2拍*根號l/g

回答 (1)

天同

2013-01-27 3:09 am

✔ 最佳答案

Length of pendulumn = L
Mass of pendulumn bob = m
Acceleration due to gravity = g

Relsove weight mg in directions perpendicular to the pendumn string, which is mg.sin(theta)

Hence, use: force = mass x acceleration
mg.sin(theta) = - m.a
g.sin(theta) = -a
multiply both sides by L
gL.sin(theta) = -L.a

If angle (theta) < 5 degrees, sin(theta) = (theta)
thus, gL(theta) = -L.a
But displacement x = L.(theta)
a.L = -g.x
a = -(g/L).x

Since (g/L) is a constant, the acceleration a is thus proportional to displacement x. The motion is "simple harmonic".
Hence, a = -(w^2).x
where w is the angular frequency, given by w^2 = g/L

But w = 2.pi/T (pi = 3.14159...)
T = (2.pi)/w = (2.pi).square-root[L/g]

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