phy SHM - Ans.fyi 參考答案

phy SHM

用戶107913

2011-06-29 8:04 pm

An object moves vertically with SHM just behind a wall
From the other side of the wall,
the object is visible in each cycle for 2s
and hidden behind the wall for 6s .

The maximum height reached by the object relative to the top of the wall is 0.3m .

Find the amplitude of the motion.

回答 (1)

天同

2011-06-30 12:00 am

✔ 最佳答案

The periof of the simple harmonic motion (SHM)
= (2 + 6) s = 8 s

Time needed for the object to move from the max height to the top of the wall = 2/2 s = 1s

Since the SHM can be desribed by the equation: x = A.cos(wt)
where x is the displacement of the object,
A is the amplitude
w is the angular frequency (= 2.pi/8 s^-1)
t is the time, with t = 0 s when the object is at max height

(A - 0.3 )= Acos[(2.pi/8) x 1]
i.e. A - 0.3 = A.cos(pi/4)
A - 0.3 = 0.707A
A = 0.3/0.293 m = 1.02 m
.

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