Pendulum clock when taken to moon

[utkarshakash]

Homework Statement

A pendulum clock that keeps correct time on the Earth is taken to the moon. It will run

a) at correct rate
b)6 times faster
c)√6 times faster
d)√6 times slower

Homework Equations

The Attempt at a Solution

$T_{earth} = 2\pi \sqrt{\dfrac{L}{g}} \\ T_{moon} = 2\pi \sqrt{\dfrac{L}{g/6}}$

Dividing i) by ii)

$\dfrac{T_{earth}}{T_{moon}} = \frac{1}{√6} \\ T_{moon} = √6T_{earth}$

This implies option c) is correct but my book says it is option d).

 

[lewando]

The period is longer so the frequency must be...

 

[sankalpmittal]

Homework Statement

A pendulum clock that keeps correct time on the Earth is taken to the moon. It will run

a) at correct rate
b)6 times faster
c)√6 times faster
d)√6 times slower

Homework Equations

The Attempt at a Solution

$T_{earth} = 2\pi \sqrt{\dfrac{L}{g}} \\ T_{moon} = 2\pi \sqrt{\dfrac{L}{g/6}}$

Dividing i) by ii)

$\dfrac{T_{earth}}{T_{moon}} = \frac{1}{√6} \\ T_{moon} = √6T_{earth}$

This implies option c) is correct but my book says it is option d).

Use concept ,

T $\alpha$ 1/√g

As √g reduces by √6 on moon , this implies time period on moon will be √6 times that of Earth , as you got. You interpreted your answer wrongly. If the time period increases , pendulum will oscillate slower or faster for a given displacement of the bob ?

 

[utkarshakash]

Use concept ,

T $\alpha$ 1/√g

As √g reduces by √6 on moon , this implies time period on moon will be √6 times that of Earth , as you got. You interpreted your answer wrongly. If the time period increases , pendulum will oscillate slower or faster for a given displacement of the bob ?

Thanks for pointing out my mistake

 

[phinds]

This implies option c) is correct but my book says it is option d).

Forget the math for a minute and just think about it logically. Would you really expect a pendulum clock when moved to lower gravity to have the pendulum swing FASTER? Really ?