The period of a simple pendulum of length L feet is given by T=2πLg−−√~seconds. We assume that g, the acceleration due to gravity on the surface of the earth, is 32 feet per second per second. If the pendulum is that of a clock that keeps good time when L=4 feet, how much time will the clock gain in 24 hours if the length of the pendulum is decreased to 3.97 feet? (Use differentials and evaluate the necessary derivative at L=4 feet.)

Respuesta :

Given that,

The acceleration due to gravity on the surface of the earth = 32 feet/s²

We need to calculate the time

Using formula of time period

[tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]

On differentiating with respect to l

[tex]\dfrac{dT}{dL}=2\pi\times\dfrac{1}{2}(\dfrac{l}{g})^{-\frac{1}{2}}\times\dfrac{1}{g}[/tex]

[tex]\dfrac{dT}{dL}=\dfrac{\pi}{g}\times(\dfrac{g}{L})^{\frac{1}{2}}[/tex]

Put the value into the formula

[tex]\dfrac{dT}{dL}=\dfrac{\pi}{32}\times(\dfrac{32}{4})^{\frac{1}{2}}[/tex]

[tex]\dfrac{dT}{dL}=0.277\ sec[/tex]

If the length of the pendulum is decreased to 3.97 feet.

We need  to calculate the time

Using formula of time period

[tex]\dfrac{dT'}{dL}=\dfrac{\pi}{g}\times(\dfrac{g}{L})^{\frac{1}{2}}[/tex]

Put the value into the formula

[tex]\dfrac{dT'}{dL}=\dfrac{\pi}{32}\times(\dfrac{32}{3.97})^{\frac{1}{2}}[/tex]

[tex]\dfrac{dT'}{dL}=0.278\ sec[/tex]

We need to calculate the gain time

Using formula for time

[tex]\dfrac{dT''}{dL}=\dfrac{dT'}{dL}-\dfrac{dT}{dL}[/tex]

Put the value into the formula

[tex]\dfrac{dT''}{dL}=0.278-0.277[/tex]

[tex]\dfrac{dT''}{dL}=0.001\ sec[/tex]

Hence, The clock gain the time in 24 hours is 0.001 sec.