Question

A simple pendulum suspended in a rocket ship has a period T0. Assume that the rocket ship is near the earth in a uniform gravitational field.

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

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

If the ship moves upward with a constant velocity, the period increases.

If the ship accelerates upward, the period increases.

If the mass of the pendulum doubles, the period

increases.

* true or false answers only

T0= t knot

Answer 1

• If the ship accelerates downward at 9.81 m/s², the pendulum will no longer oscillate.

Answer: True

As the ship accelerates downward at 9.81 m/s² near the earth's surface, it is falling freely and the acceleration becomes zero, which makes the period infinity. That means it no longer oscillates.

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

Answer: True

We know,

The length of the pendulum is doubles, l = 2l

We know,T = 2 √(l / g)

T = 2 √(2l / g)

= √2 * 2 √(l / g)

= √2 * T₀

• If the ship moves upward with a constant velocity, the period increases.

Answer: False

As the velocity of the ship remains constant , there is no acceleration.

• If the ship accelerates upward, the period increases.

Answer: False

If the ship accelerates upwards, the acceleration of the pendulum would be increased, so the period of oscillation of the pendulum decreased.

• If the mass of the pendulum doubles, the period increases.

Answer: False

The period of a pendulum depend on the length but does not depend on the mass.