The acceleration due to gravity on the surface of the moon is 1.7 m s-2. The time period of a simple pendulum on the moon if its time period on the earth is 3.5 s is (Given, g = 9.8 m s-2).

Question

The acceleration due to gravity on the surface of the moon is 1.7 m s-2. The time period of a simple pendulum on the moon if its time period on the earth is 3.5 s is (Given, g = 9.8 m s-2). (A) 2.2 s (B) 4.4 s (C) 8.4 s (D) 16.8 s

Tardigrade Admin · Accepted Answer

For moon, gm = 1.7 m s-2 For earth, ge = 9.8 m s-2, Te = 3.5 s But, Tm = 2π√(l/gm) and Te = 2π√(l/ge) ∴ (Tm/Te) = √(ge/gm) or Tm = √(ge/gm)× Te = √(9.8/1.7)× 3.5 = 8.4 s

Q. The acceleration due to gravity on the surface of the moon is $1.7 m s^{- 2}$ . The time period of a simple pendulum on the moon if its time period on the earth is $3.5 s$ is (Given, $g = 9.8 m s^{- 2})$ .

$2.2 s$

$4.4 s$

$8.4 s$

$16.8 s$

For moon, $g_{m} = 1.7 m s^{- 2}$
For earth, $g_{e} = 9.8 m s^{- 2}, T_{e} = 3.5 s$
But, $T_{m} = 2 π \frac{l}{g _{m}}$ and $T_{e} = 2 π \frac{l}{g _{e}}$
$∴ \frac{T _{m}}{T _{e}} = \frac{g _{e}}{g _{m}}$ or
$T_{m} = \frac{g _{e}}{g _{m}} \times T_{e}$
$= \frac{9.8}{1.7} \times 3.5$
$= 8.4 s$

Q. The acceleration due to gravity on the surface of the moon is 1.7ms−2. The time period of a simple pendulum on the moon if its time period on the earth is 3.5s is (Given, g=9.8ms−2).

2.2s

4.4s

8.4s

16.8s