Suppose that a grandfather clock that has a pendulum with a length of 1.5 meter. You find that the clock runs fast so that it gains 10 seconds every hour. By how much should the pendulum be lengthened? Write your answer in mm to two significant figures.

To solve this problem, we need to calculate how much the pendulum should be lengthened to make the clock run accurately.

The period of a pendulum is given by the formula:

\[
T = 2\pi \sqrt{\frac{L}{g}}
\]

Where:
- \( T \) is the period of the pendulum (time for one complete swing),
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \).

### Step 1: Understand the clock's behavior
The clock gains 10 seconds every hour, meaning its period is slightly shorter than it should be. One hour is 3600 seconds, so in a perfect clock, the period is:

\[
T_{\text{correct}} = \frac{3600}{N} \quad \text{where \( N \) is the number of oscillations per hour}.
\]

Since the clock is gaining 10 seconds every hour, its period is slightly shorter than this. The incorrect period \( T_{\text{fast}} \) is:

\[
T_{\text{fast}} = \frac{3600 - 10}{N} = \frac{3590}{N}.
\]

### Step 2: Ratio of the periods
The ratio of the fast period to the correct period is:

\[
\frac{T_{\text{fast}}}{T_{\text{correct}}} = \frac{3590}{3600}.
\]

This ratio is also related to the lengths of the pendulum before and after adjustment. The period is proportional to the square root of the length, so:

\[
\frac{T_{\text{fast}}}{T_{\text{correct}}} = \sqrt{\frac{L_{\text{fast}}}{L_{\text{correct}}}}.
\]

Squaring both sides:

\[
\left(\frac{3590}{3600}\right)^2 = \frac{L_{\text{fast}}}{L_{\text{correct}}}.
\]

### Step 3: Solving for the new length
Let \( L_{\text{correct}} \) be the current length of 1.5 meters. Solving for the new length \( L_{\text{fast}} \):

\[
L_{\text{fast}} = 1.5 \times \left(\frac{3590}{3600}\right)^2.
\]

Now, let's calculate the new length.The new length of the pendulum should be approximately \( 1.4917 \, \text{m} \).

### Step 4: Determine the length difference
Now, we calculate how much the pendulum should be lengthened:

\[
\Delta L = L_{\text{correct}} - L_{\text{fast}} = 1.5 \, \text{m} - 1.4917 \, \text{m}.
\]

We'll express this difference in millimeters.The pendulum should be lengthened by approximately \( 8.32 \, \text{mm} \) to correct the clock.

Would you like further details or have any questions?

Here are 5 related questions to explore further:
1. How would the time gained or lost change if the pendulum length was doubled?
2. What would happen to the pendulum's period if gravity were different, such as on another planet?
3. How does the amplitude of the pendulum's swing affect the period?
4. If the clock loses time instead of gaining it, how would you adjust the pendulum?
5. What are other factors that could affect the accuracy of a pendulum clock?

**Tip:** Pendulum clocks are highly sensitive to small changes in length. Even minor adjustments can significantly impact timekeeping!

Suppose that a grandfather clock has a pendulum with a length of 1.5 meters. You find that the clock runs fast so that it gains 10 seconds every hour. By how much should the pendulum be lengthened? Write your answer in mm to two significant figures.

This problem explains how to calculate the necessary pendulum length adjustment for a grandfather clock that gains 10 seconds every hour. It covers key concepts like the pendulum period formula and proportionality in pendulum mechanics. Suitable for high school students in Grades 11-12.

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