Question

A simple pendulum suspended in a rocket ship has a period _{0}" alt="T_{0}" align="absmiddle" class="latex-formula">. Assume that the rocket ship is near the earth in a uniform gravitational field.
True or false:
1. If the ship moves upward with a constant velocity, the period increases.
2. If the length of the pendulum is doubled, the new period will be: square root of 2 times $T_{0}$.
3. If the ship accelerates upward, the period increases.
4. If the ship accelerates downward at 9.81 $m/s^{2}$, the pendulum will no longer oscillate.
5. If the mass of the pendulum is halved, the period decreases.

Answer 1

(1) If the ship moves upward with a constant velocity, the period increases, false.

(2) If the length of the pendulum is doubled, the new period will be: square root of 2 times T₀. true.

(3) If the ship accelerates upward, the period increases, false

(4) If the ship accelerates downward at 9.81 m/s² , the pendulum will no longer oscillate, false.

(5) If the mass of the pendulum is halved, the period decreases, false.

Period of a simple pendulum

The period of a simple pendulum is given as;

T = 2π√(L/g)    ------- (1)

where;

• L is length of the pendulum
• g is acceleration due to gravity

When the ship moves upward with a constant velocity, the period will not change.

When the length of the pendulum is doubled

T ∝√L

when, L = 2L₀

T = T₀√2

Thus, if the length of the pendulum is doubled, the new period will be: square root of 2 times T₀.

when ship accelerates upward

t = √2h/g

where;

• g is acceleration due to gravity

The acceleration due to gravity will be constant, hence the period will not change.

when mass of the pendulum is halved

T = √I/mgd

where;

• m is mass of the pendulum

when the mass is halved, the period increases.

Learn more about period of pendulum here: brainly.com/question/26524032

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