Question

The period of oscillation of a simple pendulum of length L suspended from the roof of a rocket

accelerating upwards with a constant acceleration (g) is given by:

A. ∞

B. 0

C. 2π

q

L

2g

D. 2π

q

L

g

Answer 1

Final Answer : Time Period :
$2\pi \sqrt{ \frac{l}{2g} }$
Steps and Understanding :


1) Time Period of a simple Pendulum is given by T =
$2\pi \sqrt{ \frac{l(eff)}{g(eff)} } \\ \\ where \: g(eff) = \frac{f(exerted)}{m}$
F(exerted ) is force exerted by rocket on Bob.

2) Length of Pendulum, l= L

For Free Body Diagram of Bob, see pic

F(net) = ma
=> F(exered) -mg=mg
=> F(exerted) = 2mg

So,
g(eff) = 2g

3) Therefore, time period of Simple Pendulum is T =
$2\pi \sqrt{ \frac{l}{2g} }$