Question

A mass of $5\text{kg}$ is suspended from a spring of stiffness constant $4000\text{N/m}$. The system is fitted with a damper with a damping coefficient of $0.2$. The mass is pulled down by $50\text{mm}$ and then released. Calculate the damped frequency.

• A. $4\text{Hz}$
• B. $4.5\text{Hz}$
• C. $5\text{Hz}$
• D. $5.52\text{Hz}$

Answer 1

The correct option is B $4.5\text{Hz}$

The angular frequency of free oscillations is given by
${\omega }_0=\sqrt{\frac{k}{m}}$
From the data given in the question, we get
${\omega }_0=\sqrt{\frac{4000}{5}}=28.28\text{rad/s}$
Let ${\omega }^{\prime }$ be the angular frequency of damped oscillations.
Then,
${\omega }^{\prime }=\sqrt{{\omega }_0^2-{\gamma }^2}$
Given, $\gamma =0.2$
$\therefore {\omega }^{\prime }=\sqrt{(28.28{)}^2-(0.2{)}^2}=28.27\text{rad/s}$
$\Rightarrow f^{\prime }=\frac{{\omega }^{\prime }}{2\pi }=\frac{28.27}{2\times 3.14}\approx 4.5\text{Hz}$
Thus, option (b) is the correct answer.