Question

A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?

Answer 1

Answer:

0.99Hz

Explanation:

Using F= -mx ( spring force)

At equilibrium the gravitational force will be balanced by the spring force so mg= kx

K= mg/ 0.25 N/m

But

Frequency f= 1/2pi √g/0.25

Frequency is 0.99Hz

Answer 2

The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz

Frequency of oscillation based problem:

What information do we have?

Length starched = 2.5 cm

F = Kx

We know that

F = mg

So,

mg = Kx

K/m = g/x

$f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }$

Frequency of oscillation = 3.15 Hz

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