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A spring having with a spring constant $1200\; N m ^{-1}$ is mounted on a hortzontal table as shown in 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.
Solution
Spring constant, $k=1200\, N m ^{-1}$
Mass, $m=3\, kg$
Displacement, $A=2.0 \,cm =0.02\, cm$
Frequency of oscillation $v$, is given by the relation:
$v=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$
Where, $T$ is the time period
$\therefore v=\frac{1}{2 \times 3.14} \sqrt{\frac{1200}{3}}=3.18\, m / s$
Hence, the frequency of oscillations is $3.18 \,m / s$
Maximum acceleration $(a)$ is given by the relation:
$a=\omega^{2} \,A$
$\omega=$ Angular frequency $=\sqrt{\frac{k}{m}}$
$A=$ Maximum displacement
$\therefore a=\frac{k}{m} A=\frac{1200 \times 0.02}{3}=8\, ms ^{-2}$
Hence, the maximum acceleration of the mass is $8.0 \,m / s ^{2}$
Maximum velocity, $v_{\max }=A \omega$
$=A \sqrt{\frac{k}{m}}=0.02 \times \sqrt{\frac{1200}{3}}=0.4\, m / s$
Hence, the maximum velocity of the mass is $0.4\, m / s$
