Answer :
Answer:
46.08 feet.
Step-by-step explanation:
Given that, for length, [tex]l[/tex] feet, the time period, [tex]f(l)[/tex], of the pendulum is
[tex]f(l)=2\pi\sqrt{\frac{l}{32}}[/tex].
As the time period equals the time taken by the pendulum in one back and forth motion, so, actually, the given [tex]2.4\pi[/tex] seconds to swing back and forth is the time period.
Let [tex]l_0[/tex] feet be the length of the pendulum for the time period [tex]2.4\pi[/tex] seconds, so
[tex]f(l_0)=2.4\pi[/tex]
[tex]\Rightarrow 2\pi\sqrt{\frac{l_0}{32}}=2.4\pi[/tex]
[tex]\Rightarrow \sqrt{\frac{l_)}{32}}=1.2[/tex]
[tex]\Rightarrow \frac{l_0}{32}=(1.2)^2[/tex] [squaring on both the sides]
[tex]\Rightarrow l_0=1.44\times32[/tex]
[tex]\Rightarrow l_0=46.08[/tex]
Hence, the approximate length of a pendulum that takes [tex]2.4 \pi[/tex] seconds to swing back and forth is 46.08 feet.
