A pendulum bob on a 2 m string is displaced `60^(@)` from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path

Options

A. `sqrt2ms^(-1)`
B. `sqrt(9.8)ms^(-1)`
C. `4.43ms^(-1)`
D. `(1)/(sqrt2)ms^(-1)`

Explanation

To solve the problem of finding the speed of the pendulum bob as it passes through the lowest point in its path, we can use the principle of conservation of mechanical energy. Here are the step-by-step calculations:

### Step 1: Identify the parameters
- Length of the pendulum string (l) = 2 m
- Angle of displacement (θ) = 60 degrees
- Acceleration due to gravity (g) = 9.81 m/s² (approximately)

### Step 2: Calculate the height (h) of the bob when displaced
When the pendulum bob is displaced at an angle of 60 degrees, it rises to a certain height (h) above the lowest point. The height can be calculated using the formula:
\[ h = l - l \cos(\theta) \]
Substituting the values:
\[ h = 2 - 2 \cos(60^\circ) \]
Since \(\cos(60^\circ) = 0.5\):
\[ h = 2 - 2 \times 0.5 = 2 - 1 = 1 \text{ m} \]

### Step 3: Apply conservation of mechanical energy
At the highest point (when the bob is displaced), all the energy is potential energy (PE), and at the lowest point, all the energy is kinetic energy (KE). Thus, we can set the potential energy equal to the kinetic energy:
\[ PE = KE \]
\[ mgh = \frac{1}{2}mv^2 \]

### Step 4: Cancel mass (m) from both sides
Since mass (m) appears on both sides of the equation, we can cancel it out:
\[ gh = \frac{1}{2}v^2 \]

### Step 5: Solve for v (speed of the bob)
Rearranging the equation gives:
\[ v^2 = 2gh \]
Substituting the known values:
\[ v^2 = 2 \times 9.81 \times 1 \]
\[ v^2 = 19.62 \]
Taking the square root:
\[ v = \sqrt{19.62} \]
\[ v \approx 4.43 \text{ m/s} \]

### Conclusion
The speed of the bob as it passes through the lowest point in its path is approximately **4.43 m/s**.

---