Question

A spring is hung from the ceiling. When a block is attached to its end, it stretches $2.0\mathrm{cm}$before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?

Short answer

The frequency of the oscillation is$f=3.5Hz$.

Step-by-step solution

Given.

Given

• $x=2.0cm=0.02m$

• $g=9.8m.s^{-2}$
  

Required $f$

Finding frequency.

Using equilibrium equation of spring

$kx=mg\Rightarrow \left(1\right)$

Equation $\left(1\right)$also can be write in form

$\frac{g}{x}=\frac{k}{m}\Rightarrow \left(2\right)$

Compare equation $\left(2\right)$by angular frequency equation

${\omega }^2=(2\pi f{)}^2=\frac{k}{m}\Rightarrow (3)$

From (2) and (3) get that

${\omega }^2=(2\pi f{)}^2=\frac{g}{x}\Rightarrow f=\frac{1}{2\pi }\sqrt{\frac{g}{x}}\Rightarrow (4)$

Substitution in (4) to get that

$f=\frac{1}{2\pi }\sqrt{\frac{9.8}{0.02}}=3.5\mathrm{Hz}$