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

question 132

Multiple Choice

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

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