( \(a\) ) Prove that the change in period \(P\) of a physical pendulum with temperature is given by \(\Delta P=\frac{1}{2} \alpha P \Delta T .(b)\) A clock pendulum made of invar has a period of \(0.500 \mathrm{~s}\) and is accurate at \(20^{\circ} \mathrm{C}\). If the clock is used in a climate where the temperature averages \(30^{\circ} \mathrm{C}\), what approximate correction to the time given by the clock is necessary at the end of 30 days?

The formula for change in period of a physical pendulum with temperature is given by \( \Delta P = \frac{1}{2} \alpha P \Delta T \), where \( \Delta P \) is the change in period, \( \alpha \) is the linear thermal expansion coefficient, \( P \) is the initial period, and \( \Delta T \) is the change in temperature. This formula can be derived from the formula for the period of a physical pendulum \( P = 2\pi \sqrt{\frac{I}{mgd}} \), where \( I \) is the moment of inertia, \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( d \) is the distance from the pivot point to the center of mass. As temperature changes, \( d \) changes, but \( I \), \( m \), and \( g \) remain constant. Using the principle of linear expansion \( \Delta L = \alpha L \Delta T \), we can write \( \Delta d = \alpha d \Delta T \), which when substituted into the period equation, results in \( \Delta P = \frac{1}{2} \alpha P \Delta T \).

Using the above formula, the correction to the time for a clock pendulum of invar, having a period of 0.500 s at 20°C, when used in a climate where the temperature averages 30°C, can be calculated as follows. First, we need to determine \( \Delta T \), which is the change in temperature. This is done by subtracting the initial temperature from the final temperature, thus \( \Delta T = 30° C - 20° C = 10° C \). Invar is an alloy known for its low thermal expansion coefficient, which is approximately \( \alpha = 0.0000012°C^{-1} \). Substituting these values and the period into the formula gives \( \Delta P = \frac{1}{2} \times 0.0000012°C^{-1} \times 0.500 s \times 10°C = 0.000003 s \). This is the correction per period. In a day, there are \( 86400 \) seconds, so the number of periods in a day is \( \frac{86400 s }{ 0.500 s/period } = 172800 \) periods. Therefore, the total correction in a day is \( 0.000003 s/period \times 172800 periods/day = 0.5184 s/day \). Finally, since we need the correction at the end of 30 days, the total correction will be \( 0.5184 s/day \times 30 days = 15.552 s \).