Wave Mechanics Question 597
Question: Disregarding gravity, find the period of oscillation of the particle connected with four springs as shown in the figure. (Given: $ \theta =45^{o},\beta =30^{o} $ )
Options:
A) $ \pi \sqrt{\frac{2m}{k}} $
B) $ \sqrt{\frac{2m\pi }{k}} $
C) $ \sqrt{\frac{m\pi }{2k}} $
D) $ \pi \sqrt{\frac{m}{2k}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ 2k,{{\sin }^{2}}\theta =k _{1},or,k _{1}=2k,{{\sin }^{2}}\theta $ and $ k _{2}=2(2k){{\sin }^{2}}\beta $ Then $ k _{eq}=k _{1}+k _{2}=2k[{{\sin }^{2}}\theta +2{{\sin }^{2}}\beta ] $
$ =2k[{{\sin }^{2}}45^{o}+2{{\sin }^{2}}30^{o}]=2k( \frac{1}{2}+\frac{1}{2} )=2k $ Then $ T=2\pi \sqrt{\frac{m}{k _{eq}}}=2\pi \sqrt{\frac{m}{2k}}=\pi \sqrt{\frac{2m}{k}} $
BETA
