Question

a seconds pendulum is taken to a place where 'g' falls to 1/4th. how is time period of the pendulum affected,if at all?give reasons what will be the new time period?​

Answer 1

Time period of a pendulum is inversely proportional to the square root of acceration due to gravity.

T α 1/√g

given that gravity falls to one-fourth.
g' = g/4
let original time period = T
final time period = T'


$\frac{T' }{T} = \sqrt{ \frac{g}{g'} } \\ T' = T \sqrt{ \frac{g}{g'} } \: \\ T' = \sqrt{ \frac{g}{ \frac{g}{4} } } \\ T' = T \sqrt{4} = 2T$
So the time period becomes 2 times the original time period.

Time period of a seconds' pendulum = 2s
New Time period = 2×2s = 4s


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