TS EAMCET · Physics · Oscillations
A vertical spring mass system has the same time period as simple pendulum undergoing small oscillations. Now, both of them are put in an elevator going downwards with an acceleration \(5 \mathrm{~m} / \mathrm{s}^2\). The ratio of time period of the spring mass system to the time period of the pendulum is (Assume, acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(\sqrt{\frac{3}{2}}\)
- B \(\sqrt{\frac{2}{3}}\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
We know that time period of a spring mass system, \[ \mathrm{T}_{\mathrm{l}}=2 \pi \sqrt{\frac{m}{k}} \] where, \(m=\) mass of body and \(k=\) force constant of the spring Time period of simple pendulum, \[ \mathrm{r}_2=2 \pi \sqrt{\frac{l}{g}} \] According to the question,…
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