your grandfather clocks pendulum has a length of 0.9930 m. part a if the clock runs slow and loses 16 s per…

Explanation:

Step1: Recall the pendulum - period formula

The period of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$. In one day ($t = 86400\ s$), the number of oscillations $n$ for a correct - time - keeping pendulum with length $L_1$ and period $T_1$ is $n=\frac{t}{T_1}$, and for the slow - running pendulum with length $L_2$ and period $T_2$ is also $n=\frac{t - \Delta t}{T_2}$, where $\Delta t = 16\ s$. Since $T_1 = 2\pi\sqrt{\frac{L_1}{g}}$ and $T_2 = 2\pi\sqrt{\frac{L_2}{g}}$, and $n$ is the same in both cases, we have $\frac{t}{2\pi\sqrt{\frac{L_1}{g}}}=\frac{t-\Delta t}{2\pi\sqrt{\frac{L_2}{g}}}$. The $2\pi$ and $\sqrt{g}$ terms cancel out, giving $\frac{t}{\sqrt{L_1}}=\frac{t - \Delta t}{\sqrt{L_2}}$.

Step2: Solve for $L_2$

Cross - multiply the equation $\frac{t}{\sqrt{L_1}}=\frac{t - \Delta t}{\sqrt{L_2}}$ to get $t\sqrt{L_2}=(t - \Delta t)\sqrt{L_1}$. Then $\sqrt{L_2}=\frac{t - \Delta t}{t}\sqrt{L_1}$. Square both sides: $L_2=\left(\frac{t - \Delta t}{t}\right)^2L_1$. Given $t = 86400\ s$, $\Delta t = 16\ s$, and $L_1 = 0.9930\ m$. First, calculate $\frac{t - \Delta t}{t}=\frac{86400 - 16}{86400}=\frac{86384}{86400}$. Then $L_2=\left(\frac{86384}{86400}\right)^2\times0.9930$. $\left(\frac{86384}{86400}\right)^2=\left(1-\frac{16}{86400}\right)^2\approx1 - 2\times\frac{16}{86400}$ (using the approximation $(1 - x)^2\approx1 - 2x$ for $x\ll1$). $L_2\approx\left(1-\frac{32}{86400}\right)\times0.9930=0.9930-\frac{32\times0.9930}{86400}$. $\frac{32\times0.9930}{86400}\approx\frac{32\times1}{86400}=\frac{32}{86400}\approx0.00037$. So $L_2\approx0.9930 - 0.00037=0.99263\ m$.

Step3: Calculate the change in length

The change in length $\Delta L=L_2 - L_1$. $\Delta L=0.99263 - 0.9930=- 0.00037\ m\approx - 0.37\ mm$. Rounding to two significant figures, $\Delta L=-0.37\ mm$.

Answer:

• 0.37 mm