Wave Mechanics Question 578
Question: A particle at the end of a spring executes S.H.M with a period $ t _{1} $ , while the corresponding period for another spring is $ t _{2} $ . If the period of oscillation with the two springs in series is T then
Options:
A) $ {{T}^{-1}}=t _{1}^{-1}+t _{2}^{-1} $
B) $ T^{2}=t _{1}^{2}+t _{2}^{2} $
C) $ T=t _{1}+t _{2} $
D) $ {{T}^{-2}}=t _{1}^{-2}+t _{2}^{-2} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ t _{1}=2\pi \sqrt{\frac{m}{k _{1}},},t _{2}=2\pi \sqrt{\frac{m}{k _{2}}} $ when springs are in series then, $ k _{eff}=\frac{k _{1}k _{2}}{k _{1}+k _{2}} $
$ \
Therefore ,T=2\pi \sqrt{\frac{m(k _{1}+k _{2})}{k _{1}k _{2}}} $
$ \
Therefore ,T=2\pi \sqrt{\frac{m}{k _{2}}+\frac{m}{k _{1}}}=2\pi \sqrt{\frac{t _{2}^{2}}{{{(2\pi )}^{2}}}+\frac{t _{1}^{2}}{{{(2\pi )}^{2}}}} $
$ \Rightarrow ,T^{2}=t _{1}^{2}+t _{2}^{2} $
BETA
