91Ó°ÊÓ

The mass of the head is $m=\left(15\mathrm{g}\right)\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)=0.015\mathrm{kg}$

The frequency of the oscillation of the head is $f=4.0Hz$

The amplitude of oscillation of the head at time $t=4.0s$is$A(4.0s)=0.5cm$

In part a) The objective is to determine

the spring constant of the spring on which the head

is mounted

In part b) The objective is to find out the head’s damping constant

The frequency of oscillation of spring can be found form first equation

$f=\frac{1}{2\mathrm{Ï€}}\sqrt{\frac{k}{m}}$

Rearrange and solve for $k$

$f^2=\frac{1}{4{\mathrm{Ï€}}^2}\frac{k}{m}$

$k=4{\mathrm{Ï€}}^{2}m{f}^2$

Substituting values

$k=4{\mathrm{Ï€}}^2\left(0.015\mathrm{kg}\right)(4.0\mathrm{Hz}{)}^2$

$=9.47N/m$

According equation $\left(1\right)$, the amplitude of the head's oscillation is expressed as function of time as follows,

$A\left(t\right)=A_0{e}^{−bt/2m}$

$\frac{A\left(t\right)}{A_0}=e^{−bt/2m}$

$\ln \left[\frac{A\left(t\right)}{A_0}\right]=−\frac{bt}{2m}$

$b=−\frac{2m}{t}\ln \left[\frac{A\left(t\right)}{A_0}\right]$

Substitute numerical value in it

$b=−\frac{2\left(0.015\mathrm{kg}\right)}{4.0\mathrm{s}}\ln \left(\frac{0.5\mathrm{cm}}{3.0\mathrm{cm}}\right)$

$=0.013kg/s$