al phy 08 Q9

wing

2010-12-31 7:35 pm

In order to check the speed of a camera shutter,thew camera was used to photogragh the bob of a simple pendulum swinging in front of a horizontal scale.the extreme positions of the bob were at the 450mm and 550mm marks onthe scale.The photograph showed that the bob moved from the 500 mm marks on the scale.The photograph showed that the bob moved from the 500 mark to the 525 mark when the shutter was open.
If the period of the pendulum was1.5 s ,the shutter remained opened for approximatelu
A1/2s
b.1/4 s
c.1/8 s
D. 3/8 s
The answers is C
Would you please explain why the anser is C?

回答 (1)

天同

2010-12-31 8:09 pm

✔ 最佳答案

Since 450 mm and 550 mm are the two extreme positions, the equilibrium position is at (450+550)/2 mm = 500 mm. The amplitude of oscillation is thus equal to (550-500) mm = 50 mm

Given the period of oscillation T = 1.5 s, hence the angular frequency w is
w = 2.pi/1.5 s^-1, where pi = 3,14159...

The oscillating pendulum follows the equation:
x = A.sin(wt)
where x is the displacement at time t,
A is the amplitude (= 50 mm)
w is the angular frequency (= 2.pi/1.5 s^-1)

Thus, (525 - 500) = 50[sin(2.pi/1.5)t]
i.e. sin[(2.pi/1.5)t] = 1/2
(2.pi/1.5)t = arc-sin(0.5) = 30 x pi/180
t = (1.5/2.pi) x (30 x pi/180) s = 0.125 s

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