A simple pendulum is executing S.H.M. around point \(\mathrm{O}\) between the end points \(B\) and \(C\) with a periodic time of \(6 \mathrm{~s}\). If the distance between \(\mathrm{B}\) and \(\mathrm{C}\) is \(20 \mathrm{~cm}\) then in what time will the bob move from \(C\) to \(D\) ? Point \(D\) is at the mid-point of \(C\) and \(\mathrm{O}\). (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)

For the pendulum performing S.H.M., the total distance from point B to point C is 20 cm. To find the distance between point C and point D, we must determine the mid-point between C and O. The distance between C and D is half the distance between C and O. Since the distance between B and C is 20 cm, we can say that the distance between O and B, or O and C is half of this value, which is 10 cm. Therefore, the distance between point C and point D is \[C\!D = \frac{1}{2}(C\!O) = \frac{1}{2}(10~\!cm) = 5~\!cm\]

In this step, we need to find angular displacement from C to D. Since it's a pendulum, so we can use the formula \[\theta = \sin^{-1} \frac{x}{L}\] where \(\theta\) is the angular displacement, \(x\) is the linear displacement and \(L\) is the length of the pendulum from point O to the either B or C. Here, we have \(x = C\!D = 5~\!cm\). We need to find \(L\), the length of the pendulum, which can be calculated by dividing \(B\!C\) by \(2 \sin \theta\). Since \(B\!C = 20\ cm\), \[ L = \frac{B\!C}{2 \sin{\theta}} = \frac{20 \mathrm{~cm}}{2 \sin{\theta}} = \frac{10 \mathrm{~cm}}{\sin{\theta}}\] Now, plugging L and x in the formula, we get \[\theta = \sin^{-1} \frac{5 \mathrm{~cm}}{\frac{10 \mathrm{~cm}}{\sin{\theta}}}\] After simplification, we find \[\theta = \sin^{-1} \frac{1}{2} = \sin^{-1} \left(\frac{1}{2} \right) = 30^{\circ} ~(\text{or} ~\frac{\pi}{6} \text{radians})\] So, the angular displacement from C to D is \(\frac{\pi}{6}\) radians.

Here, the simple pendulum is performing S.H.M., and it is mentioned that its periodic time is 6 seconds. The relation between the angular displacement and time for a simple pendulum is given by the formula: \[T(t) = \frac{2 \pi \sqrt{\frac{L}{g} }}{\theta}\] Where \(T(t)\) is the time it takes for the object to move through the angle \(\theta\), \(L\) is the length of the pendulum, and \(g\) is the acceleration due to gravity. Now, the periodic time for this pendulum is 6 seconds. From our previous calculation of angular displacement, the pendulum moves through an angle of \(\frac{\pi}{3}\) radians between B and C. Therefore, we can use this information to determine the time it takes to go through \(\frac{\pi}{6}\) radians (from C to D): \[T(t) = \frac{6}{2} = 3 \mathrm{s}\]

After calculating the time it takes for the pendulum to travel from point C to point D, we get it as 3 seconds. So now, we can match our answer to the options given in the problem: (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\) The correct answer is (C) \(3 \mathrm{~s}\).