A clock pendulum made of invar has a period of 0.5 s at `20^(@)C`. If the clock is used in a climate where average temperature is `30^(@)C`,what correction may be necessary at the end of 30 days `alpha_("invar")=7xx10^(-7)(.^(@)C)^(-1)`

Answer

To solve the problem, we need to calculate the correction in the period of a clock pendulum made of invar due to a temperature change from 20°C to 30°C. The following steps will guide us through the calculation:

### Step 1: Understand the relationship between period and length
The period \( T \) of a pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity. Since \( g \) is constant, any change in the period will be due to a change in the length \( L \).

### Step 2: Determine the change in length due to temperature
The change in length \( \Delta L \) of the pendulum due to a change in temperature \( \Delta T \) can be calculated using the formula:
\[ \Delta L = L \alpha \Delta T \]
where \( \alpha \) is the coefficient of linear expansion of invar, and \( \Delta T \) is the change in temperature.

### Step 3: Calculate the temperature change
Given:
- Initial temperature \( T_1 = 20°C \)
- Final temperature \( T_2 = 30°C \)

The change in temperature is:
\[ \Delta T = T_2 - T_1 = 30°C - 20°C = 10°C \]

### Step 4: Substitute values into the length change formula
Substituting the values into the length change formula:
\[ \Delta L = L \cdot (7 \times 10^{-7} \, \text{°C}^{-1}) \cdot (10 \, \text{°C}) = 7 \times 10^{-6} L \]

### Step 5: Relate change in length to change in period
Using the relationship between the change in period \( \Delta T \) and the change in length:
\[ \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta L}{L} \]
Substituting \( \Delta L \):
\[ \frac{\Delta T}{T} = \frac{1}{2} \cdot \frac{7 \times 10^{-6} L}{L} = \frac{7 \times 10^{-6}}{2} = 3.5 \times 10^{-6} \]

### Step 6: Calculate the change in period
The original period \( T \) is given as \( 0.5 \, \text{s} \):
\[ \Delta T = T \cdot \frac{\Delta T}{T} = 0.5 \, \text{s} \cdot 3.5 \times 10^{-6} = 1.75 \times 10^{-6} \, \text{s} \]

### Step 7: Calculate the total time for 30 days
Now, we need to calculate the total correction over 30 days. The number of seconds in 30 days is:
\[ 30 \, \text{days} = 30 \times 86400 \, \text{s} = 2592000 \, \text{s} \]

### Step 8: Calculate the total correction over 30 days
The total correction over 30 days is:
\[ \text{Total Correction} = \Delta T \times \text{Number of periods in 30 days} \]
The number of periods in 30 days is:
\[ \text{Number of periods} = \frac{2592000 \, \text{s}}{0.5 \, \text{s}} = 5184000 \]
Thus,
\[ \text{Total Correction} = 1.75 \times 10^{-6} \, \text{s} \times 5184000 \approx 9.072 \, \text{s} \]

### Step 9: Final correction calculation
Since the correction is calculated per period, we divide by 2 (as derived earlier):
\[ \text{Final Correction} = \frac{9.072}{2} \approx 4.536 \, \text{s} \]

### Conclusion
The necessary correction at the end of 30 days is approximately **4.536 seconds**.

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