Question

The time, in seconds, that it takes a pendulum to swing back and forth is modeled by the equation below. f (l) = 2 pi StartRoot StartFraction l Over 32 EndFraction EndRoot, where l is the length of the pendulum in feet What is the approximate length of a pendulum that takes 2.4 pi seconds to swing back and forth?

Answer 1

Answer:

46.08 feet.

Step-by-step explanation:

Given that, for length, $l$ feet, the time period, $f(l)$, of the pendulum is

$f(l)=2\pi\sqrt{\frac{l}{32}}$.

As the time period equals the time taken by the pendulum in one back and forth motion, so, actually, the given $2.4\pi$ seconds to swing back and forth is the time period.

Let $l_0$ feet be the length of the pendulum for the time period $2.4\pi$ seconds, so

$f(l_0)=2.4\pi$

$\Rightarrow 2\pi\sqrt{\frac{l_0}{32}}=2.4\pi$

$\Rightarrow \sqrt{\frac{l_)}{32}}=1.2$

$\Rightarrow \frac{l_0}{32}=(1.2)^2$ [squaring on both the sides]

$\Rightarrow l_0=1.44\times32$

$\Rightarrow l_0=46.08$

Hence, the approximate length of a pendulum that takes $2.4 \pi$ seconds to swing back and forth is 46.08 feet.