Question

A) For the next scene, an astronaut travels to a moon of Planet W. She needs to measure the local gravity on the planet. She decides to use a 1.5 m long pendulum. She finds it takes her pendulum 6.09 seconds to complete one full oscillation. What is the acceleration due to gravity on this moon?
a. 1.6 m/s²
b.2.1 m/s²
c. 2.3 m/s²
d.1.9 m/s²

B) For the next scene, an astronaut travels to a moon of Planet W. She needs to measure the local gravity on the planet. She decides to use a 1.5 m long pendulum. She finds it takes her pendulum 6.09 seconds to complete one full oscillation. What is the frequency of this pendulum?
a. 0.41 Hz
b. 1.2 Hz
c. 0.28 Hz
d. 0.16 Hz

C) If this pendulum were brought back to Earth to shoot the remaining scenes of the movie, how would its frequency on Earth compare to its frequency while on the moon?
a. 1.2 Hz
b. 0.16 Hz
c. 0.41 Hz
d. 0.28 Hz

D) The director is worried about moving the pendulum from Planet W’s moon to the earth. He is worried that a change in the mass of the pendulum bob (which is necessary for the script) will change the frequency of oscillation. Calculate the change in frequency if the mass is twice as heavy as the original.
a. 100% increase in frequency.
b. 50% increase in frequency.
c. 50% decrease in frequency.
d. No change in frequency.
e. 100% decrease in frequency.

Answer 1

A)

L = length of the pendulum = 1.5 m

T = time period of oscillation of pendulum = 6.09 s

g = acceleration due to gravity

Time period of the pendulum is given as

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

6.09 = (2 x 3.14) sqrt(1.5/g)

g = 1.6 m/s²

a. 1.6 m/s²

B)

f = frequency

frequency is given as

f = 1/T

f = 1/6.09

f = 0.16 Hz

d. 0.16 Hz

C)

g_e = acceleration due to gravity on earth = 9.8 m/s²

g_m = acceleration due to gravity on moon = 1.6 m/s²

f_e = frequency on earth

f_m = frequency on moon = 0.16 Hz

frequency is given as

f = (1/(2$\pi$)) sqrt(g/L)

hence

f_e / f_m = sqrt(g_e/g_m)

f_e /0.16 = sqrt(9.8/1.6)

f_e = 0.41 Hz

c. 0.41 Hz

D)

frequency is given as

f = (1/(2$\pi$)) sqrt(g/L)

the frequency does not depend on mass, hence

d. No change in frequency.