A simple pendulum executing SHM with a period of 6 s between two extreme positions B and C about a point O. if the length of the arc BC is 10 cm, how long will the pendulum take the move from position C to a position D towards O exactly midway between C and O?

Options

A. 0.5 s
B. 1 s
C. 1.5 s
D. 3 s

Explanation

To solve the problem, we need to determine the time it takes for a simple pendulum to move from position C to position D, which is exactly midway between position O (the mean position) and position C. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the given information
- The period (T) of the pendulum is 6 seconds.
- The length of the arc BC is 10 cm.
- The mean position O is the midpoint of the arc BC.

### Step 2: Calculate the amplitude of the pendulum
The amplitude (A) is half the length of the arc BC. Therefore:
\[ A = \frac{BC}{2} = \frac{10 \text{ cm}}{2} = 5 \text{ cm} \]

### Step 3: Determine the position of point D
Point D is midway between O and C. Since the distance from O to C is equal to the amplitude (5 cm), the distance from O to D is:
\[ OD = \frac{OC}{2} = \frac{5 \text{ cm}}{2} = 2.5 \text{ cm} \]

### Step 4: Use the formula for displacement in SHM
The displacement (x) at any time (t) can be expressed as:
\[ x = A \cos(\omega t) \]
where \(\omega\) is the angular frequency given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting the period:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \]

### Step 5: Set up the equation for position D
At position D, the displacement (x) is 2.5 cm. Therefore, we can write:
\[ 2.5 = 5 \cos\left(\frac{\pi}{3} t\right) \]
Dividing both sides by 5 gives:
\[ \frac{1}{2} = \cos\left(\frac{\pi}{3} t\right) \]

### Step 6: Solve for \(\frac{\pi}{3} t\)
The cosine of \(\frac{\pi}{3}\) is \(\frac{1}{2}\), so:
\[ \frac{\pi}{3} t = \frac{\pi}{3} \]

### Step 7: Solve for time (t)
To find t, we can cancel \(\pi\) from both sides:
\[ t = 1 \text{ second} \]

### Conclusion
The time taken for the pendulum to move from position C to position D is **1 second**.

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