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Solve the Problem =11πsec= 11 \pi \mathrm { sec } , Frequency

Question 127

Multiple Choice

Solve the problem.
-Determine the period and frequency of oscillation when a pendulum of length 11 feet is released after being displaced 2 radians. Round constants to 8 decimal places, if necessary.


A) Period =11πsec= 11 \pi \mathrm { sec } , frequency =111π= \frac { 1 } { 11 \pi } cycles per sec
B) Period =2π2.90909091= 2 \pi \sqrt { 2.90909091 } sec, frequency =12π0.34375= \frac { 1 } { 2 \pi } \sqrt { 0.34375 } cycles per sec
C) Period =2π0.34375= 2 \pi \sqrt { 0.34375 } sec, frequency =12π2.90909091= \frac { 1 } { 2 \pi } \sqrt { 2.90909091 } cycles per sec
D) Period =π11= \frac { \pi } { 11 } sec, frequency =11π= \frac { 11 } { \pi } cycles per sec

Correct Answer:

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