(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.
<h3>Period of a simple pendulum</h3>
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.
<h3>When the length of the pendulum is doubled</h3>
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.
<h3>when ship accelerates upward</h3>
t = √2h/g
where;
- g is acceleration due to gravity
The acceleration due to gravity will be constant, hence the period will not change.
<h3>when mass of the pendulum is halved</h3>
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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