PHY SHM - Ans.fyi 參考答案

PHY SHM

用戶107913

2011-06-30 1:10 am

A small block of mass of 0.1kg is suspended from the ceiling by a light spring
of force constant 12 Nm^-1 . If the block is projected vertically downwards with
a speed of 0.5 ms^-1 from its equilibrium position .
What is the maximum acceleration of the block in its subsequent motion ?

回答 (1)

天同

2011-06-30 4:29 am

✔ 最佳答案

The block will subsequently perform simple harmonic motion (SHM) about the equilibrium position.

In a SHM, the max acceleration occurs at the extreme positions, i.e. at the amplitudes.

Hence, max acceleration a = (w^2)A
where w is the nagular frequency of oscillation
A is the amplitude of the oscillation

In a spring-mass oscillation system, w = k/m
where k is the spring constant, and m is the mass of the block
thus, w = 12/0.1 s^-1 = 120 s^-1

IN a SHM, the max velocity V occurs at the equilibrium position, which is given by
V = w.A
thus, 0.5 = 120A
A = 0.5/120 m = 4.167 x 10^-3 m

Therefore, max acceleration a = (120)^2 x (4.167 x 10^-3) m/s2 = 60 m/s2

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