Solve the problem.
-Determine the period and frequency of oscillation when a pendulum of length 7 feet is released after being displaced 2 radians.
A) Period $= 2 \pi \sqrt { 4.57142857 }$ sec, frequency $= \frac { 1 } { 2 \pi } \sqrt { 0.21875 }$ cycles per sec
B) Period $= 7 \pi \sec$, frequency $= \frac { 1 } { 7 \pi }$ cycles per sec
C) Period $= 2 \pi \sqrt { 0.21875 } \mathrm { sec }$, frequency $= \frac { 1 } { 2 \pi } \sqrt { 4.57142857 }$ cycles per sec
D) Period $= \frac { \pi } { 7 } \mathrm { sec }$, frequency $= \frac { 7 } { \pi }$ cycles per sec

Question

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). Converting the length to meters (if necessary) is not required here as we're working in feet and using the appropriate value for $g$ in feet. The frequency is the reciprocal of the period. Given $l = 7 \, \text{feet}$, the period $T = 2\pi\sqrt{\frac{7}{32}}$ matches the form in choice C when simplified, and the frequency is correctly given as the reciprocal of the period.

Solve the Problem =2π4.57142857= 2 \pi \sqrt { 4.57142857 }=2π4.57142857​ Sec, Frequency

Solve the Problem $= 2 \pi \sqrt { 4.57142857 }$ Sec, Frequency