Applied Maths - Mechanics 45

還不明白的薑

2009-08-16 5:46 am

Refer to the diadram : http://img11.imageshack.us/img11/1221/shm.jpg

A weight of mass M is attached to a smooth, weightless pulley system equipped with light inelastic strings and light springs with elastic constants k1 and k2 as known in the figure. Show that, if the weight M is displaced slightly downwards from its equilibrium position and released, it will execute simple harmonic motion. Find the period of the oscillation.

回答 (1)

用戶9112

2009-08-16 8:52 am

✔ 最佳答案

Referring to the diagram:
http://img151.imageshack.us/img151/1590/pulleys.jpg.
In the figure, if z is changed by an amount x, then y will be changed by an amount x/2.
If w is changed by an amount u, the t will be changed by an amount 2u.
Since the pulley system is light,
T1 = 2T2 ; T2 = 2T3
δT1 = k1u from the equilibrium value.
The extension in k2 will be changed by an amount = y – t = x/2 – 2u
δT2 = k2(x/2 – 2u)
δT1 = 2δT2 => k1u = 2k2(x/2 – 2u)
k1u + 4k2u = k2x
u = k2x/(k1 + 4k2)
δT1 = k1u = k1k2x/(k1 + 4k2)
δT2 = δT1/2 = k1k2x/[2(k1 + 4k2)]
δT3 = δT2/2 = k1k2x/[4(k1 + 4k2)]
This δT3 is acting in the opposite direction of the displacement.
Therefore Md2x/dt2 + k1k2x/[4(k1 + 4k2)] = 0
d2x/dt2 + k1k2x/[4M(k1 + 4k2)] = 0 => SHM
Period = 2π√[4M(k1 + 4k2)]/(k1k2)

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