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

Options

A. `sqrt2 m//s`
B. `sqrt 9.8 m//s`
C. `4.43 m//s`
D. `1//sqrt2 m//s`

Explanation

To find the speed of the pendulum bob as it passes through the lowest point in its path, we can use the principle of conservation of energy or the work-energy theorem. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Length of the string (L) = 2 m
- Angle of displacement (θ) = 60°
- Initial speed (u) = 0 m/s (since the bob is released from rest)

### Step 2: Calculate the height (h) from which the bob is released
When the bob is displaced by an angle of 60°, the height (h) can be calculated using the formula:
\[ h = L - L \cos(\theta) \]
Substituting the values:
\[ h = 2 - 2 \cos(60°) \]
Since \(\cos(60°) = \frac{1}{2}\):
\[ h = 2 - 2 \cdot \frac{1}{2} = 2 - 1 = 1 \text{ m} \]

### Step 3: Apply the work-energy theorem
According to the work-energy theorem:
\[ \text{Work done by gravity} = \Delta KE \]
The work done by gravity (W) is given by:
\[ W = mgh \]
Where:
- \( m \) = mass of the bob (we will see it cancels out)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( h \) = height calculated above (1 m)

The change in kinetic energy (ΔKE) is:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2 \]

### Step 4: Set up the equation
Equating the work done by gravity to the change in kinetic energy:
\[ mgh = \frac{1}{2}mv^2 \]

### Step 5: Cancel mass (m) from both sides
Since mass \( m \) appears on both sides, we can cancel it out:
\[ gh = \frac{1}{2}v^2 \]

### Step 6: Rearrange to solve for speed (v)
Rearranging the equation gives:
\[ v^2 = 2gh \]
Taking the square root:
\[ v = \sqrt{2gh} \]

### Step 7: Substitute the values
Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( h = 1 \, \text{m} \):
\[ v = \sqrt{2 \cdot 9.8 \cdot 1} \]
\[ v = \sqrt{19.6} \]
\[ v \approx 4.427 \, \text{m/s} \]

### Conclusion
The speed of the bob as it passes through the lowest point in its path is approximately **4.427 m/s**.

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