NEET · Physics · STD 11 - 13. oscillations
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is \(\frac{x}{2}\) times its original time period. Then the value of \(x\) is:
- A \(\sqrt{2}\)
- B \(2 \sqrt{3}\)
- C \(4\)
- D \(\sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(T^{\prime}=2 \pi \sqrt{\frac{\ell^{\prime}}{g}} \text { where } \ell^{\prime}=\frac{\ell}{2}\) \(T=2 \pi \sqrt{\frac{\ell}{g}}\)
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