Question

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

Answer 1

Answer:

The correct option is C: 0.31 s.

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:

$T = 2\pi \sqrt{\frac{m}{k}}$    (2)

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

$T = 2\pi \sqrt{\frac{m}{\frac{mg}{\Delta y}}}$

$T = 2\pi \sqrt{\frac{\Delta y}{g}}$

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:

$T = 2\pi \sqrt{\frac{0.025 m}{10 m/s^{2}}} = 0.31 s$                   

Hence, the oscillation period is 0.31 s.

The correct option is C: 0.31s.

I hope it helps you!