Question

A spring having with a spring constant 1200 N m${}^{-1}$ is mounted on a horizontal table as shown in the Figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Answer 1

Spring constant, $k=1200Ncal{m}^{-1}$

Mass, $m=3kg$

Displacement, $A=2.0cm=0.02m$

(i) Frequency of oscillation

v, is given by the relation:

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

where, T is time period

$\therefore v=\frac{1}{2\times 3.14}\sqrt{\frac{1200}{3}}=3.18Hz$

Hence, the frequency of oscillations is 3.18 cycles per second.

(ii) Maximum acceleration (a) is given by the relation:

$a={\omega }^{2}A$
where,

$\omega =$ Angular frequency $=\sqrt{\frac{k}{m}}$

$A$ = maximum displacement

$\therefore a=\frac{k}{m}A=\frac{1200\times 0.02}{3}=8m{s}^{-2}$

Hence, the maximum acceleration of the mass is $8.0m/s^2$

(iii) Maximum velocity, $v_{max}=A\omega$

$=A\sqrt{\frac{k}{m}}=0.02\times \sqrt{\frac{1200}{3}}=0.4m/s$

Hence, the maximum velocity of the mass is 0.4 m/s.