Question

A spring having a spring constant $1200{\text{Nm}}^{-1}$ is mounted on a horizontal table as shown. A mass of $3.0\text{kg}$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $0.2\text{cm}$ and released. Determine
(i) The frequency of oscillations.
(ii) The maximum acceleration of the mass, and
(iii) the maximum speed of the mass?

• A. $3.18\text{Hz},8{\text{m/s}}^2,0.40\text{m/s}$
• B. $1.57\text{Hz},16{\text{m/s}}^2,0.80\text{m/s}$
• C. $3.18\text{Hz},32{\text{m/s}}^2,1.6\text{m/s}$.
• D. $3.18\text{Hz},16{\text{m/s}}^2,0.80\text{m/s}$

Answer 1

The correct option is A $3.18\text{Hz},8{\text{m/s}}^2,0.40\text{m/s}$

Given that,
Spring constant $k=1200{\text{Nm}}^{-1}$
mass $m=3\text{kg}$
The distance at which mass is pulled from its equilibrium $x=0.2\text{cm}$
(i) Frequency,
$f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
$=\frac{1}{2\times 3.142}\times \sqrt{\frac{1200}{3}}$
$f=3.18\text{Hz}$
(ii) The acceleration is given by
$a=-{\omega }^{2}x=-\frac{k}{m}x$
or $|a_{max}|=\frac{k}{m}|x_{max}|$
i.e., acceleration will be maximum, when $x$ is maximum.
i.e., $x_{max}=A=0.02m$
$\therefore a=\frac{1200}{3}\times 0.02=8.0{\text{ms}}^{-2}$
(ii) The maximum speed of the mass is given by
$v=A\omega =A\sqrt{\frac{k}{m}}=0.02\times \sqrt{\frac{1200}{3}}$
$=0.40{\text{ms}}^{-1}$