Question

the mass and diameter of a planet are two times those of earth. if a seconds pendulum is taken to it, what is the time period of the pendulum in seconds?

Answer 1

METHOD I:

Time period of a second’s pendulum on earth is 2 s.

On earth, T = 2π√(L/g)

=> L = g/π2

We know,

g = GM/R2

So for the planet, acceleration due to gravity will be, g/ = G(2M)/(2R)2 = GM/(2R2) = g/2

So, time period of the second’s pendulum on that planet will be,

T/ = 2π√(L/g/) = 2π√[(g/π2)/(g/2)] = 2√2 s

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Or, METHOD II (Image):

Answer 2

Time period of Pendulum is $2\sqrt2$ seconds

Explanation:

Time period formula is ,

$T = 2\pi \sqrt{\frac {l}{g}}$

It is given as time period 2 second on earth's surface.

As you know,

Acceleration due to gravity on any planet is,

$g_p = \frac {GM_p}{(R_p)^2}$

where $M_p$ is Mass of planet And $R_p$ is Radius of Planet.

So as the Mass and diameter are two times so you can see ,

$g_p = \frac {g}{2}$

Now put the value of g in the time period.

You will get, Time period on planet will be,

$T_p = \sqrt 2T$

Where T= Time on earth= 2 second.

So time period on the planet will be $2\sqrt2$ Second.