Question

Show that the motion of a simple pendulum is simple harmonic and hence
derive an equation for its time period. The mass and radius of a planet
are double that of the earth. If the time period of a simple pendulum on
the earth is T, find the time period on the planet.​

Answer 1

Answer:

Time period of the simple pendulum is given as

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

The new time period of the pendulum is given as

$T' = \sqrt{\sqrt2}T$

Explanation:

As we know that the restoring force on the pendulum is given as

$F = mg sin\theta$

so we have

$a = g sin\theta$

$a = - g\frac{x}{L}$

now we have

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

so motion of simple pendulum is SHM

Now as the radius and mass of the planet is double then acceleration due to gravity is given as

$g_p = \sqrt{\frac{G(2M)}{(2R)^2}}$

$g_p = \frac{g}{\sqrt2}$

now the new time period is given as

$T' = 2\pi\sqrt{\frac{L}{g/\sqrt2}$

so it is given as

$T' = \sqrt{\sqrt2}T$

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Topic : SHM of pendulum

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Answer 2

Explanation:

Time period of the simple pendulum is given as

T = 2\pi\sqrt{\frac{L}{g}}T=2πgL

The new time period of the pendulum is given as

T' = \sqrt{\sqrt2}TT′=2T

Explanation:

As we know that the restoring force on the pendulum is given as

F = mg sin\thetaF=mgsinθ

so we have

a = g sin\thetaa=gsinθ

a = - g\frac{x}{L}a=−gLx

now we have

T = 2\pi\sqrt{\frac{L}{g}}T=2πgL

so motion of simple pendulum is SHM

Now as the radius and mass of the planet is double then acceleration due to gravity is given as

g_p = \sqrt{\frac{G(2M)}{(2R)^2}}gp=(2R)2G(2M)

g_p = \frac{g}{\sqrt2}gp=2g

now the new time period is given as

T' = 2\pi\sqrt{\frac{L}{g/\sqrt2}

so it is given as

T' = \sqrt{\sqrt2}TT′=2T