Solve the problem.
-Determine the period and frequency of oscillation when a pendulum of length 9 feet is released after being displaced 2 radians. Round constants to 8 decimal places, if necessary.
A) Period $= 2 \pi \sqrt { 3.55555556 } \mathrm { sec }$, frequency $= \frac { 1 } { 2 \pi } \sqrt { 0.28125 }$ cycles per sec
B) Period $= 9 \pi \mathrm { sec }$, frequency $= \frac { 1 } { 9 \pi }$ cycles per sec
C) Period $= \frac { \pi } { 9 }$ sec, frequency $= \frac { 9 } { \pi }$ cycles per sec
D) Period $= 2 \pi \sqrt { 0.28125 }$ sec, frequency $= \frac { 1 } { 2 \pi } \sqrt { 3.55555556 }$ cycles per sec

Question

Solve the problem.
-Determine the period and frequency of oscillation when a pendulum of length 9 feet is released after being displaced 2 radians. Round constants to 8 decimal places, if necessary. 
A) Period $= 2 \pi \sqrt { 3.55555556 } \mathrm { sec }$, frequency $= \frac { 1 } { 2 \pi } \sqrt { 0.28125 }$ cycles per sec
B) Period $= 9 \pi \mathrm { sec }$, frequency $= \frac { 1 } { 9 \pi }$ cycles per sec
C) Period $= \frac { \pi } { 9 }$ sec, frequency $= \frac { 9 } { \pi }$ cycles per sec
D) Period $= 2 \pi \sqrt { 0.28125 }$ sec, frequency $= \frac { 1 } { 2 \pi } \sqrt { 3.55555556 }$ cycles per sec

Quizplus · Accepted Answer

The period of a pendulum is given by $T = 2\pi\sqrt{\frac{l}{g}}$, where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$ or $32 \, \text{ft/s}^2$ for calculations in feet). For a pendulum of length 9 feet, $T = 2\pi\sqrt{\frac{9}{32}}$, which simplifies to $2\pi\sqrt{0.28125}$. The frequency is the reciprocal of the period, hence $f = \frac{1}{T} = \frac{1}{2\pi\sqrt{0.28125}}$.

Solve the Problem =2π3.55555556sec= 2 \pi \sqrt { 3.55555556 } \mathrm { sec }=2π3.55555556​sec

Solve the Problem $= 2 \pi \sqrt { 3.55555556 } \mathrm { sec }$