Question

A large pendulum with a 200-lb gold plated bob 12 inches in diameter is on display in the lobby of the united nations building. the pendulum has a length of 75 ft. it is used to show the rotation of the earth—for this reason it is referred to as a foucault pendulum. part a if the pendulum were to be taken to the moon, where the acceleration of gravity is g/6, would the period increase, decrease, or stay the same?

Answer 1

The period of one full swing depends on the length of the pendulum and on gravity. The period of each full swing would be longer on the moon, with less gravity.

The rotation of the plane of the swings doesn't depend on the length of the string OR on gravity. It only depends on the latitude of the place where the pendulum hangs, and the rotation period of the body it's located on.

On Earth, it's (24 hours)/(sine of latitude).

On the moon, it would be (27.32 days)/(sine of latitude).

Answer 2

The time period of oscillation of the pendulum will increase on the surface of the moon.

Further Explanation:

The time period of oscillation of a pendulum is given by:

$\boxed{T=2\pi\sqrt{\frac{l}{g}}}$

Here, $T$  is the time period of oscillation of the pendulum, $l$  is the length of the pendulum and $g$  is the acceleration due to gravity on the surface.

The above expression of the time period of simple pendulum shows that the time period of oscillation is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity on the surface.

$\boxed{T \propto \sqrt l }$

$\boxed{T \propto \frac{1}{{\sqrt g }}}$

So, when the pendulum is taken to the surface of the moon, the length of the pendulum remains constant. Therefore, the variation in the time period of oscillation of pendulum depends on the acceleration due to gravity.

Since the value of acceleration due to gravity decreases on the surface of the moon, there will be a significant increase in the time period of oscillation of the pendulum.

The length of the pendulum is $75\,{\text{ft}}$  .

$\begin{aligned}l&=75{\text{ft}}\times\left({\frac{{1\,{\text{m}}}}{{3.281\,{\text{ft}}}}}\right)\\&=22.86\,{\text{m}}\\\end{aligned}$

Time period of oscillation on Earth:

$\begin{aligned}{T_{earth}}&=2\pi\sqrt{\frac{{22.86}}{{9.8}}}\\&=2\pi\times 1.527\\&=9.6\,{\text{s}}\\\end{aligned}$

Time period of oscillation on moon:

$\begin{aligned}{T_{moon}}&=2\pi\sqrt{\frac{{22.86}}{{\left({\frac{{9.8}}{6}}\right)}}}\\&=2\pi\times 3.74\,{\text{s}}\\&=2{\text{3}}{\text{.5}}\,{\text{s}}\\\end{aligned}$

Therefore, the time period of oscillation of the pendulum will increase on the surface of the moon.

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Answer Details:

Grade: College

Subject: Physics

Chapter: Simple Pendulum

Keywords:

Pendulum, united nations building, length 75ft, 200-lb gold plated, Foucault pendulum, time period, increases, moon, acceleration due to gravity, g/6.