Question

please help me answers all part of question 11 correctly.

11. 1.25/10 points Previous Answers SHM: Pendulum 1/6 Submissions Used e Content My Notes A Part 1: Pendulum on allen planet You are on a mysterious planet and notice that a 2.7 m long simple pendulum has a period of 3s when its angular amplitude is for less A) Determine the value for 3.7974 m/s the free full acceleration on the surface of this planet .) If you DOUBLED the length of the pendulum, what would its period be? 7 6070 Part 2 Pendulum on Earth You are on Earth and notice that the pendulum in grandfather clock is 1.57 m long. Determine the following Witsanger amplitude is 2

Answer 1

Given Data

T = Time period = 5.3 s

L = length of the pendulum = 2.7 m

g = acceleration due to gravity on the planet

Solution

A)

Time period of the pendulum is given as

T = 2$\pi$ sqrt(L/g)

5.3 = (2 x 3.14) sqrt(2.7/g)

0.712 = 2.7/g

g = 3.79 m/s²

B)

L' = new length of the pendulum = L/2 = 2.7/2 = 1.35 m

new time period of the pendulum is given as

T' = 2$\pi$ sqrt(L'/g)

T' = (2 x 3.14) sqrt(1.35 / 3.79)

T' = 3.748 sec  

Given Data

length l = 1.57 m

angular amplitude = theta = 2.2 degrees = 0.03839 rad

time period T = 2*3.14*sqrt(l/g)

T = 2*3.14*sqrt(1.57/9.8) = 2.514 sec --> Answer for T

frequency f = 1/T = 1/2.514 = 0.398 sec^-1 = 0.398 Hz --> Answer for frequency, f

angular frequency w = 2pif = 2*3.14*0.398 = 2.498 rad/s --> Answer for w

amplitude A = theta*l = 0.03839*1.57 = 0.0602 m --> Answer for A

maximum velocity v max = Aw = 0.0602*2.498 = 0.151 m/s ---> Answer for v max

maximum acceleration = A max = Aw^2 = 0.0602*2.498^2 = 0.3762 m/s^2 --> Answer for A max