The length of a pendulum is measured as 20.0 cm. The time interval for 100 oscillations is measured as 90 s with a stop watch of 1 s resolution. Find the approximate percentage change in g.

Answer

To solve the problem step by step, we will follow the outlined method to find the approximate percentage change in \( g \).

### Step 1: Understand the formula for the time period of a pendulum
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( T \) is the time period,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 2: Rearranging the formula to find \( g \)
Squaring both sides gives:
\[ T^2 = 4\pi^2 \frac{L}{g} \]
Rearranging this to solve for \( g \):
\[ g = \frac{4\pi^2 L}{T^2} \]

### Step 3: Calculate the values of \( L \) and \( T \)
From the problem:
- Length of the pendulum \( L = 20.0 \, \text{cm} = 0.20 \, \text{m} \)
- Time interval for 100 oscillations \( T_{100} = 90 \, \text{s} \)
- Therefore, the time period \( T \) for one oscillation is:
\[ T = \frac{T_{100}}{100} = \frac{90}{100} = 0.90 \, \text{s} \]

### Step 4: Calculate \( g \) using the values of \( L \) and \( T \)
Substituting the values into the formula for \( g \):
\[ g = \frac{4\pi^2 (0.20)}{(0.90)^2} \]

Calculating \( \pi^2 \):
\[ \pi^2 \approx 9.87 \]
Now substituting:
\[ g = \frac{4 \times 9.87 \times 0.20}{0.81} \approx \frac{7.896}{0.81} \approx 9.73 \, \text{m/s}^2 \]

### Step 5: Calculate the uncertainties in \( L \) and \( T \)
- The uncertainty in length \( \Delta L = 0.1 \, \text{cm} = 0.001 \, \text{m} \)
- The uncertainty in time \( \Delta T = 1 \, \text{s} \)

### Step 6: Calculate relative errors
The relative error in \( g \) can be expressed as:
\[ \frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T} \]

Calculating each term:
1. For \( \frac{\Delta L}{L} \):
\[ \frac{\Delta L}{L} = \frac{0.001}{0.20} = 0.005 \]

2. For \( \frac{\Delta T}{T} \):
\[ \frac{\Delta T}{T} = \frac{1}{90} \approx 0.0111 \]

Now substituting these values into the relative error formula:
\[ \frac{\Delta g}{g} = 0.005 + 2 \times 0.0111 \approx 0.005 + 0.0222 = 0.0272 \]

### Step 7: Convert relative error to percentage error
To find the percentage change in \( g \):
\[ \text{Percentage change} = \frac{\Delta g}{g} \times 100 \approx 0.0272 \times 100 \approx 2.72\% \]

### Final Answer
The approximate percentage change in \( g \) is about **2.72%**, which can be rounded to **3%**.

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