KCET · Physics · Oscillations
A simple pendulum is suspended from the ceiling of a lift. When the lift is at rest its time period is With what acceleration should the lift be accelerated upwards in order to reduce its period to \(\mathrm{T} / 2\) ? ( is acceleration due to gravity).
- A 2
- B 3
- C 4
- D 6
Answer & Solution
Correct Answer
(B) 3
Step-by-step Solution
Detailed explanation
Time period of simple pendulum is given by
\(\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{~g}}}\)
When the lift is moving up with an acceleration , then time period becomes
\(T^{\prime}=2 \pi \sqrt{\frac{1}{g+a}}\)
Here, \(\mathrm{T}^{\prime}=\frac{\mathrm{T}}{2}\)
\(\Rightarrow \frac{\mathrm{T}}{2}=2 \pi \sqrt{\frac{1}{\mathrm{~g}+\mathrm{a}}}\)
Dividing Eq. (ii) by Eq. (i), we get
\(a=3 g\)
\(\mathrm{T}=2 \pi \sqrt{\frac{1}{\mathrm{~g}}}\)
When the lift is moving up with an acceleration , then time period becomes
\(T^{\prime}=2 \pi \sqrt{\frac{1}{g+a}}\)
Here, \(\mathrm{T}^{\prime}=\frac{\mathrm{T}}{2}\)
\(\Rightarrow \frac{\mathrm{T}}{2}=2 \pi \sqrt{\frac{1}{\mathrm{~g}+\mathrm{a}}}\)
Dividing Eq. (ii) by Eq. (i), we get
\(a=3 g\)
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