Question

A second's pendulum is taken to a planet where the acceleration due to gravity is 16 times that on earth will stop what would be the time period of the pendulum on the planet? *
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Answer 1

Given that, a second's pendulum is taken to a planet where the acceleration due to gravity is 16 times that on earth.

We have to find the time period of the pendulum on the planet.

We know that time period for simple pendulum is:

T = 2π √(l/g) ............(1st equation)

For second's pendulum time period is taken as 2s and acceleration due to gravity is 16 times that on earth.

⇒ acceleration due to gravity (g) = 16g

So,

Let's denote the time period for planet by T' whereas T is the time period for earth.

T' = 2π √(l/16g)

T' = 1/4 × 2π √(l/g) .........(2nd equation)

From (1st equation) we can write the (2md equation) like:

T' = 1/4 × T ........(3rd equation)

As, 2π √(l/g) = T

According to question, we are talking about the second's pendulum and as said above too that time period for second's pendulum is 2s.

Substitute value of T = 2s in (3rd equation)

T' = 1/4 × 2

T' = 2/4

T' = 1/2

T' = 0.5

Therefore, the time period of the pendulum on the planet is 0.5 sec.

Answer 2

$\text{we know that the at Earth ,the Time period is}$

$T = 2π\sqrt{\dfrac{l}{g}} [Equation (1)]$

$\text{Time period of seconds pendulum = 2s}$

$\text{Let the time period at}$

$\Rightarrow Earth = T_{1}$

$\Rightarrow Planet (x) = T_{2}$

$\text{The time period at the another planet}$

$T = 2π\sqrt{\dfrac{l}{16g}} [Equation (2)$

$\text{Dividing the first equation and second equation ,we get}$

$\dfrac{T_{1}}{T_{2}} = \dfrac{2π\sqrt{\dfrac{l}{g}}}{2π\sqrt{\dfrac{l}{16g}}}$

$\dfrac{T_{1}}{T_{2}} = \sqrt{16}$

$\dfrac{T_{1}}{T_{2}} =\dfrac{1}{4}$

$From\:the\:equation ,\dfrac{T_{1}}{T_{2}} =\dfrac{1}{4}$

$we\:can\:find\:the\:value\:of\:T_{2}$

$\Rightarrow T_{2} = \dfrac{\dfrac{1}{4}}{T_{1}}</p><p>$

$Substitute ,T_{1} for\:2$

$\Rightarrow T_{2} = \dfrac{\dfrac{1}{4}}{2}$

$\Rightarrow T_{2} =\dfrac{2}{4}</p><p>$

$\Rightarrow T_{2} = 0.5$

$\text{Hence,The time period at that planet is 0.5sec}$