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Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsOscillations5 Marks
Question
A spring having with a spring constant $1200N m^{-1}$ is mounted on a horizontal table as shown in Fig. A mass of $3kg$ is attached to the free end of the spring. The mass is then pulled sideways to a distance of $2.0cm$ and released.
Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.
✓

Answer

Spring constant, $k = 1200N m^{-1}$
Mass, m = 3kg
Displacement, A = 2.0cm = 0.02cm
  1. Frequency of oscillation v, is given by the relation:
$\upsilon=\frac{1}{\text{T}}=\frac{1}{2\pi}\sqrt{\frac{\text{k}}{\text{m}}}$
where, T is time period
$\therefore\ \upsilon=\frac{1}{2\times3.14}\sqrt{\frac{1200}{3}}$
= 3.18m/s
Hence, the frequency of oscillations is 3.18 cycles per second.
  1. Maximum acceleration (a) is given by the relation:
$\text{a}=\omega^2\text{A}$
where,
$\omega=$ Angular frequency $=\sqrt{\frac{\text{k}}{\text{m}}}$
A = maximum displacement
$\therefore\ \text{a}=\frac{\text{k}}{\text{m}}\text{A}=\frac{1200\times0.02}{3}=8\text{ ms}^{-2}$
Hence, the maximum acceleration of the mass is $8.0m/s^2$.
Maximum velocity, $\text{v}_\text{max}=\text{A}\omega$
$=\text{A}\sqrt{\frac{\text{k}}{\text{m}}}=0.02\times\sqrt{\frac{1200}{3}}=0.4\text{ m/s}$
Hence, the maximum velocity of the mass is 0.4m/s.

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