Question

A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point where the spring is not stretched. Then the mass is released and begins to oscillate. The lowest point in the oscillation is 18 cm below the point where the mass was released. What is the oscillation frequency?

Answer 1

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

$E_s=\dfrac{1}{2}kx^2$

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

$E_P=mgx$

Equating these two energies,

$\dfrac{1}{2}kx^2=mgx$

$\dfrac{k}{m}=\dfrac{2g}{x}$.............(1)

The expression for the oscillation frequency is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}$ (from equation (1))

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}$

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.