Question

4. A massless spring hangs from the ceiling, and a mass is hung from the bottom of it. The mass is supported

so that initially the tension in the spring is zero. The mass is then suddenly released. At the bottom of
its trajectory, the mass is 5 centimeters from its original position. Find its oscillation period.
(A) 0.05 s
(B) 0.07 s
(C) 0.31 s
(D) 0.44 s
(E) Not enough information is given.

Answer 1

The correct option is C: 0.31 s.

Step-by-step explanation:

When the mass is then suddenly released we have:

`F = k\Delta y`

Where:

F is the force

k: is the spring constant

Δy: is the spring displacement

Since the tension in the spring is zero, the force is the weight:

`F = mg`

Where:

m is the mass of the object

g is the gravity

`mg = k\Delta y` (1)

The oscillation period of the spring is given by:

`T = 2\pi \sqrt{(m)/(k)}` (2)

By solving equation (1) for "k" and entering into equation (2) we have:

`T = 2\pi \sqrt{(m)/((mg)/(\Delta y))}`

`T = 2\pi \sqrt{(\Delta y)/(g)}`

Since the spring will osclliates in a position between the initial position (when it is at rest) and the final position (when the mass is released and reaches the bottom), we have Δy = 2.5 cm = 0.025 m:

`T = 2\pi \sqrt{(0.025 m)/(10 m/s^(2))} = 0.31 s`

Hence, the oscillation period is 0.31 s.

The correct option is C: 0.31s.

I hope it helps you!