The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0 degree to the vertical.
(a)With what frequency does it vibrate? Assume the bob executes SHM.
(b)What is the pendulum bobs speed when it passes through the lowest point of the swing?
(c)What is the total energy stored in this oscillation, assuming no losses?
simple pendulum
2009-12-06 12:38 am
回答 (1)
2009-12-06 3:05 am
✔ 最佳答案
a. Given the bob executes S.H.M., the period of the bob can be found by the equation:T = 2pi sqrt(L / g)
T = 2pi sqrt(0.760 / 9.8)
Period, T = 1.75 s
Frequency, f = 1 / T = 1/1.75 = 0.57 Hz
b. By the law of conservation of energy
G.P.E. lost = K.E. gained
mgL(1 - cos@) = 1/2 mv^2
(9.8)(0.760)(1 - cos12*) = 1/2 v^2
Speed at the lowest point, v = 0.57 ms^-1
c. Total energy stored
= K.E. at the lowest point
= 1/2 mv^2
= 1/2 (0.365)(0.57)^2
= 0.0594 J
參考: Physics king
收錄日期: 2021-04-13 18:52:58
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