The small-angle approximation assumes that the sine of the angle can be approximated by the angle itself (in radians). For a simple pendulum, the period $T$ is given by:

$T = 2\pi \sqrt{\frac{L}{g}}$

where:

• $L = 1.5 \, \text{m}$ (length of the pendulum),
• $g = 9.803 \, \text{m/s}^2$ (acceleration due to gravity).

Step 1: Plugging in the values

$T = 2\pi \sqrt{\frac{1.5}{9.803}}$

Step 2: Compute the term inside the square root

$\frac{1.5}{9.803} \approx 0.153 \, \text{s}^2$

Step 3: Take the square root

$\sqrt{0.153} \approx 0.391 \, \text{s}$

Step 4: Multiply by $2\pi$

$T = 2\pi \cdot 0.391 \approx 2.458 \, \text{s}$

Final Answer:

$T = 2.458 \, \text{s}$

The correct answer is: T = 2.458 s

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Would you like further explanation on this derivation or how the small-angle approximation affects the period? Here are related questions you might find useful:

1. How does the small-angle approximation simplify the pendulum equation?
2. What is the effect of larger initial angles on the period of a pendulum?
3. How would the period change if the length $L$ were doubled?
4. Why does the acceleration due to gravity $g$ affect the period?
5. What are the limitations of the simple pendulum model in real-world scenarios?

Tip: Always use radians for small-angle approximations in pendulum calculations! $10^\circ$ should be converted to radians if needed, but it doesn't directly affect the formula for $T$.