Physics 205A Quiz 6, fall semester 2018

Cuesta College, San Luis Obispo, CA

the longest pendulum currently in existence is at the Oregon Convention Center in Portland, OR, a 40 kg mass swinging on a cable with a period of 9.25 seconds.[*] At that location the magnitude of the acceleration due to gravity

*g* is 9.82611 m/s

^{2}.[**]

In San Luis Obispo, CA, the magnitude of the acceleration due to gravity

*g* is 9.80844 m/s

^{2}.[***]

If a similar pendulum were constructed in San Luis Obispo, CA, and set into the motion with the same amplitude, it would have a period __________ the period of the pendulum in Portland, OR.

(A) less than.

(B) equal to.

(C) greater than.

(D) (Not enough information is given.)

[*] Martin Beech,

*The Pendulum Paradigm: Variations on a Theme and the Measure of Heaven and Earth*, BrownWalker Press (2014), p. 42.

[**]

wolframalpha.com/input/?i=gravitational+acceleration+portland,+or.

[***]

wolframalpha.com/input/?i=gravitational+acceleration+san+luis+obispo,+ca.

Correct answer (highlight to unhide):

(C)
The period of a pendulum is given by:

T = 2·π·√(

*L*/

*g*),

which does

*not* depend on the mass, and only on the length

*L* of the cable (which is the same for both locations), and the gravitational acceleration constant

*g* (which is different for these locations).

Since San Luis Obispo, CA has a slightly smaller gravitational acceleration constant (9.80844 m/s

^{2}) compared to Portland, OR (9.82611 m/s

^{2}), then the pendulum in San Luis Obispo, CA will have a slightly longer period compared to Portland, OR.

This can be shown quantitatively by writing out the pendulum period equations for both locations:

T

_{Portland} = 2·π·√(

*L*_{Portland}/

*g*_{Portland}),

T

_{SLO} = 2·π·√(

*L*_{SLO}/

*g*_{SLO}).

Since the cable length would be the same for both locations, then:

*L*_{SLO} =

*L*_{Portland},

and solving for the period for the pendulum in San Luis Obispo, CA:

*g*_{SLO}·(T

_{SLO}/(2⋅π)

^{2} =

*g*_{Portland}·(T

_{Portland}/(2⋅π)

^{2},

*g*_{SLO}·(T

_{SLO}/(

~~2⋅π~~)

^{2} =

*g*_{Portland}·(T

_{Portland}/(

~~2⋅π~~)

^{2},

(T

_{SLO})

^{2} = (T

_{Portland})

^{2}⋅(

*g*_{Portland}/

*g*_{SLO}) ,

T

_{SLO} = T

_{Portland}√(

*g*_{Portland}/

*g*_{SLO}),

T

_{SLO} = (9.25 s)·√((9.82611 m/s

^{2})/(9.80844 m/s

^{2})) = 9.2583282333... s,

which to three significant figures is 9.26 s, and thus longer than the period of the pendulum in Portland, OR.

Sections 70854, 70855

Exam code: quiz06POr7

(A) : 10 students

(B) : 10 students

(C) : 32 students

(D) : 0 students

Success level: 61%

Discrimination index (

Aubrecht & Aubrecht, 1983): 0.58