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To solve this problem, we’ll use dimensional analysis to determine the formula for the period \( t \) of a simple pendulum, involving the mass \( m \), string length \( l \), and gravitational acceleration \( g \).

### Step-by-Step Solution

1. **Identify the Dimensions of Each Quantity:**

   - **Period \( t \)**: The time for one complete swing, so its dimension is \([T]\).
   - **Mass \( m \)**: Mass, with dimension \([M]\).
   - **Length \( l \)**: The length of the pendulum, with dimension \([L]\).
   - **Gravitational Acceleration \( g \)**: This is acceleration due to gravity, with dimension \([L][T]^{-2}\).

2. **Formulate the Dimensional Equation:**

   We assume the period \( t \) depends on \( m \), \( l \), and \( g \) in some way. Thus, we can express \( t \) as:
   \[
   t = k \cdot m^a \cdot l^b \cdot g^c
   \]
   where \( k \) is a dimensionless constant, and \( a \), \( b \), and \( c \) are exponents to be determined.

3. **Substitute Dimensions and Set Up Equations:**

   Replacing each quantity with its dimensional form, we get:
   \[
   [T] = [M]^a \cdot [L]^b \cdot [L][T]^{-2}^c
   \]
   Expanding this:
   \[
   [T] = [M]^a \cdot [L]^{b+c} \cdot [T]^{-2c}
   \]

4. **Equate Dimensions:**

   Now, we match the powers of each fundamental dimension on both sides of the equation:

   - For \( M \): \( a = 0 \)
   - For \( L \): \( b + c = 0 \)
   - For \( T \): \( -2c = 1 \)

5. **Solve the System of Equations:**

   From \( a = 0 \), we see that mass \( m \) does not affect the period, so \( m \) can be disregarded.

   Solving \( -2c = 1 \) gives \( c = -\frac{1}{2} \).

   Substitute \( c = -\frac{1}{2} \) into \( b + c = 0 \):
   \[
   b - \frac{1}{2} = 0 \Rightarrow b = \frac{1}{2}
   \]

6. **Construct the Formula:**

   Substituting the values of \( a \), \( b \), and \( c \) back, we get:
   \[
   t = k \cdot l^{1/2} \cdot g^{-1/2}
   \]
   Simplifying further:
   \[
   t = k \cdot \frac{\sqrt{l}}{\sqrt{g}} = k \cdot \sqrt{\frac{l}{g}}
   \]

7. **Conclusion:**

   The formula for the period \( t \) of a simple pendulum, based on dimensional analysis, is:
   \[
   t = k \cdot \sqrt{\frac{l}{g}}
   \]
   where \( k \) is a dimensionless constant, often determined experimentally to be close to \( 2\pi \) for a simple pendulum.

Would you like more details on any part of this solution, or have any questions about how we approached the dimensional analysis?

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### Additional Questions to Explore:
1. How does the mass \( m \) affect the period of a simple pendulum if at all?
2. Why does gravitational acceleration \( g \) appear in the denominator in the formula?
3. What is the significance of the constant \( k \), and how is it determined?
4. How would the formula change if this were a different type of pendulum (e.g., a physical pendulum)?
5. Can we derive the period for other oscillatory systems using similar dimensional analysis?

#### Tip:
Dimensional analysis is a powerful tool to predict relationships between physical quantities without knowing the exact equations.

You are testing a decorative clock, to possibly be manufactured by your Consumer Electronics division, and attach a mass (m) to a string of length (l) to form a simple pendulum. Assuming that the acceleration due to gravity (g) of the earth may have an influence on the period (t) of the pendulum swing, use dimensional analysis to find a formula for t which could possibly involve m, l, and g.

Learn how to use dimensional analysis to derive a formula for the period (t) of a simple pendulum, considering the influences of mass (m), length (l), and gravitational acceleration (g). This solution explains step-by-step how to find the formula t = k * sqrt(l / g), suitable for undergraduate physics students.

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