The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth?

Question

The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth? (A) √2 second (B) 2 √2 seconds (C) (1/√2) second (D) (1/2 √2) second

Tardigrade Admin · Accepted Answer

g=(G M/R2) g prime=(G(2 M)/(2 R)2) so g prime=(g/2) T prime=2 π √(2 ℓ/g) ⇒ T prime=√2 T ⇒ T prime=2 √2sec.

Q. The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth?

$2$ second

$22$ seconds

$\frac{1}{2}$ second

$\frac{1}{2 2}$ second

$g = \frac{GM}{R ^{2}}$ & $g^{'} = \frac{G ( 2 M )}{( 2 R ) ^{2}}$
so $g^{'} = \frac{g}{2}$
$T^{'} = 2 π \frac{2 ℓ}{g}$
$\Rightarrow T^{'} = 2 T$
$\Rightarrow T^{'} = 22$ sec.

Q. The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on earth?

2​ second

22​ seconds

2​1​ second

22​1​ second

g=R2GM​ &g′=(2R)2G(2M)​ so g′=2g​ T′=2πg2ℓ​​ ⇒T′=2​T ⇒T′=22​sec.

$2$ second

$22$ seconds

$\frac{1}{2}$ second

$\frac{1}{2 2}$ second

$g = \frac{GM}{R ^{2}}$ & $g^{'} = \frac{G ( 2 M )}{( 2 R ) ^{2}}$
so $g^{'} = \frac{g}{2}$
$T^{'} = 2 π \frac{2 ℓ}{g}$
$\Rightarrow T^{'} = 2 T$
$\Rightarrow T^{'} = 22$ sec.