Question

The $15g$head of a bobble-head doll oscillates in SHM at a frequency of $4.0Hz$.

a. What is the spring constant of the spring on which the head is mounted?

b. The amplitude of the head’s oscillations decreases to $0.5cm$in $4.0s$. What is the head’s damping constant?

Short answer

The spring constant of the spring on which the head

is mounted is$k=9.47N/m$

The head’s damping constant is$b=0.013kg/s$

Step-by-step solution

Given data

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$A(4.0s)=0.5cm$

Step 2:  Required data

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

 solution Part a)

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

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

Rearrange and solve for $k$

$f^2=\frac{1}{4{\pi }^2}\frac{k}{m}$

$k=4{\pi }^{2}m{f}^2$

Substituting values

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

$=9.47N/m$

part b)

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

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

$=0.013kg/s$