[单选题]
A particle performs simple
pb vk35ym .x u2u um((buj0n,aharmonic
4 loj.a; dlt/c-bde*xmotion with a frequency of $0.05Hz$ The distance traveled by the particle in one complete oscillation is $16\,cm$ Which of the following graphs shows a correct variation of its position?
A.
B.
C.
D.
参考答案:
A
本题详细解析:
Explanation:
Since frequency is 0.05Hz,T=$\frac{1}{f}$=$\frac{1}{0.005}$=20s
The amplitude is 4cm on either side of the equilibrium position because in one full cycle it travels four amplitudes
($\frac{16}{4}=4cm$)
