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Home » A spring having with a spring constant is mounted on a horizontal table as shown in Fig. 14.24. A mass of is attached to the free end of the spring. The mass is then pulled sideways to a distance of and released.
A spring having with a spring constant 1200 \mathrm{~N} \mathrm{~m}^{-1} is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3 \mathrm{~kg} is attached to the free end of the spring. The mass is then pulled sideways to a distance of \mathbf{2}, \mathbf{0} \mathbf{~ c m} and released.
CBSE | Class 11 | NCERT | Oscillations | Physics
A spring having with a spring constant 1200 \mathrm{~N} \mathrm{~m}^{-1} is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3 \mathrm{~kg} is attached to the free end of the spring. The mass is then pulled sideways to a distance of \mathbf{2}, \mathbf{0} \mathbf{~ c m} and released.

Determine<br>(i) the frequency of oscillations,

(ii) maximum acceleration of the mass

Solution:

Spring constant is given as \mathrm{k}=1200 \mathrm{~N} \mathrm{~m}^{-1}

Mass is given as \mathrm{m}=3 \mathrm{~kg}
Displacement is given as \mathrm{A}=2.0 \mathrm{~cm}

=0.02 \mathrm{~m}

(i) Frequency of oscillation ‘ v ‘ can be calculated as,

\begin{array}{l} v=(1 / T) \\ =(1 / 2 \pi)(\sqrt{k} / \mathrm{m}) \end{array}

Where,
\mathrm{T}= time period

So,

v=\{1 /(2 \times 3.14)\} \sqrt{1200 / 3}

=3.18 \mathrm{~m} / \mathrm{s}

As a result, the frequency of oscillations is 3.18 cycles per second.

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

\mathrm{a}=\omega^{2} \mathrm{~A}

Where,

\omega=\text { Angular frequency }=\sqrt{k} / \mathrm{m}

A= Maximum displacement

So,

\begin{array}{l} a=(k / m) A \\ a=(1200 \times 0.02) / 3 \end{array}
=8 \mathrm{~m} \mathrm{~s}^{-2}

As a result, the maximum acceleration of the mass is 8.0 \mathrm{~m} / \mathrm{s}^{2}

i

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