Calculation of true thickness of the bed

GENERAL

1. Moon's gravity is about 1/6 of that on earth, what is the period of a seconds pendulum on the Moon? ( Neglect thermal contraction/ expension)

A)2*√6  sec         B) 2*√3 sec          C) 2 sec     D) 3 sec

Solution:

A second pendulum on Earth has a period of 2 seconds

The period of simple pendulum :

T= 2*π *√(L/g)  -----------------------(1)

L- length of the pendulum (m)

g- acceleration due to gravity (gm/cc)

T- time period (sec)

The time period of simple pendulum (Earth) T_E = 2*π *√(L/g_E) ------------(2)

The time period of  simple pendulum ( Moon) T_m =  2*π *√(L/g_m)-----------(3)

Dividing eq(2) by eq(3)

T_E/ T_m = ( 2*π*√ (L/g_E) ) /  ( 2*π √(L/g_m) )

T_E/ T_m = (√ (1/g_E)  /  √( 1/g_m) )

From above problem,

Suppose, Earth acceleration due to gravity is =1

The moom’s acceleration due  to gravity is = 1/6

T_E/ T_m = ( √( 1/1 ) /  √( 1/ (1/6) )

T_E/ T_m = ( 1/√6)

 T_m = √6 T_E

Therefore, T_E is 2 seconds

T_m= 2*√6

The period of the second pendulum on the moon is  =2*√6