A uniform meter stick is freely pivoted about the 0.20-m mark. If it is allowed to swing in a vertical plane with a small amplitude and friction, what is the frequency of its oscillations?
0.66 Hz
0.55 Hz
1.3 Hz
1.1 Hz
0.92 Hz
Physics Help. Answer please?
2015-12-10 4:56 pm
回答 (1)
2015-12-10 10:51 pm
✔ 最佳答案
equivalent length Leq = I / mdBy the parallel axis theorem,
I = Icm + md² = mL²/12 + m(0.3L)² = ML²(1/12 + 0.09) = 0.173ML²
so Leq = 0.173ML² / M*0.3L = 0.578L = 0.578 * 1.0m = 0.578 m
T = 2π√(Leq / g) = 2π√(0.578m / 9.8m/s²) = 1.53 s
and
f = 1 / T = 0.66 Hz ◄
Hope this helps!
收錄日期: 2021-04-21 15:39:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20151210085616AAfTLFd