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TS EAMCET · Physics · Oscillations

The time period of a simple pendulum is \(T\). When the length is increased by \(10 \mathrm{~cm}\), its period is \(T_1\). When the length is decreased by \(10 \mathrm{~cm}\), its period is \(T_2\). Then, relation between \(T, T_1\) and \(T_2\) is

  1. A \(\frac{2}{T_2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}\)
  2. B \(\frac{2}{T_2}=\frac{1}{T_1^2}-\frac{1}{T_2^2}\)
  3. C \(2 T^2=T_1^2+T_2^2\)
  4. D \(2 T^2=T_1^2-T_2^2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2 T^2=T_1^2+T_2^2\)

Step-by-step Solution

Detailed explanation

\(T^2=4 \pi^2\left(\frac{l}{g}\right)\)...(i) \(T_1^2=4 \pi^2\left(\frac{l+10}{g}\right)\)...(ii) \(T_2^2=4 \pi^2\left(\frac{l-10}{g}\right)\)....(iii) Adding Eqs. (ii) and (iii), we get…
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