MHT CET · Physics · Oscillations
A pendulum performs S.H.M. with period \(\sqrt{3}\) second in a stationery lift. If lift moves up with acceleration \(\frac{\mathrm{g}}{3}\), the period of pendulum is \([\mathrm{g}=\) acceleration due to gravity \(]\)
- A \(2 \cdot 00\) second
- B \(1 \cdot 5\) second
- C \(2 \cdot 5\) second
- D \(1 \cdot 75\) second
Answer & Solution
Correct Answer
(B) \(1 \cdot 5\) second
Step-by-step Solution
Detailed explanation
\(\mathrm{T}=\sqrt{3} \quad \mathrm{a}=\frac{\mathrm{g}}{3}\)
\(\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}+\frac{\mathrm{g}}{3}}}\)
\(\therefore \frac{T^{\prime}}{T}=\sqrt{\frac{g}{4 g / 3}}=\frac{\sqrt{3}}{2}\)
\(\mathrm{T}^{\prime}=\sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}=1.5 \mathrm{~sec}\)
\(\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}+\frac{\mathrm{g}}{3}}}\)
\(\therefore \frac{T^{\prime}}{T}=\sqrt{\frac{g}{4 g / 3}}=\frac{\sqrt{3}}{2}\)
\(\mathrm{T}^{\prime}=\sqrt{3} \times \frac{\sqrt{3}}{2}=\frac{3}{2}=1.5 \mathrm{~sec}\)
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