(i) What is the length of the pendulum of a clock which beats seconds, given that the acceleration due to gravity at a place is equal to `9.8ms^(-2)`.
(ii) Find the time period of another pendulum of length equal to half of it at the same place.
(iii) A second pendulum is taken into a mine of depth 640 m and oscillated. What is its time period?
(iv) The length of a seconds pendulum at a place is 1.02 m. Find the time period of another pendulum of length 0.51 m at the same place.

Explanation

The period of simple pendulum is given by
`T=2pisqrt(l/g)implies` The length of the pendulum
`=1=(gT^(2))/(4pi^(2))implies1=(9.8xx2^(2))/(4xx(3.14)^(2))`
`l=(9.8xx2^(2))/(4xx9.86)=0.994m~=1m`
(ii) The period of a simple pendulum is given by
`T=2pisqrt(l/g)`........1 `" ":.g=4pi^(2)l/(T^(2))`
i.e.`l/(T^(2))=g/((4p)^(2))=` constant at a given place (2)
`(l_(1))/(T_(1)^(2))=(l_(2))/(T_(2)^(2))implies1.0/(2^(2))=0.50/(T_(2)^(2))`or `T_(2)^(2)=0.50/1.01xx4=2`
`:.T_(2)=sqrt(2)=1.414s`
(iii) The time period of a simple pendulum is
`T=2pisqrt(l/g)` When a seconds pendulum is taken into a mine g changes (l constant)
The ratio of its time periods is `(T_(1))/(T_(2))=sqrt((g_(2))/(g_(1)))`
On the surface of the earth `g_(1)=g` and in the mine
`g_(2)=g(1-d/R)=(T_(1))/(T_(2))=sqrt((g(1-d/R))/g)"or"(T_(1))/(T_(2))=sqrt(1-d/R)`
Here `T_(1)=2s,d=640m=0.640km,`
`R=` Radius of the earth `=6400km,T_(2)=?`
`2/(T_(2))=sqrt(1-0.640/6400)=sqrt(1-0.0001)`
`:.T_(2)=2/(sqrt(1-0.0001))=2.0001s`
(iv) The perido of a simple pendulum is given by
`T=2pisqrt(l/g)`..........1
i.e `l/(T^(2))=g/(4pi^(2))=` constant
at a give place .....2
If `T_(1)` and `T_(2)` are the time periods of the pendulum for length `l_(1)` and `l_(2)` From eq. (2) we can write
`(l_(1))/(T_(1)^(2))=(l_(2))/(T_(2)^(2))`.....3
Here `T_(1)=2s`, when `l_(1)=1.02m`
when `l_(2)=0.51m,T_(2)=?`
Using eq. 3 `1.02/(2^(2))=0.51/(T_(2)^(2))` or `T_(2)^(2)=0.51/1.02xx4=2`
`:.T_(2)=sqrt(2)=1.414s`. When the length increasses the time period of the pendulum increases. So the clock loses time or runs slow.