91Ó°ÊÓ

The amplitude of the motion can be obtained by using the concept of total energy, which is equal to the sum of kinetic energy and spring potential energy.

Given data:

The mass of an object is \(m = 2.7\;{\rm{kg}}\).

The spring constant is \(k = 310\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}\).

The distance of an object from the equilibrium position is \(x = 0.020\;{\rm{m}}\).

The speed of an object is \(v = 0.55\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).

(a)

At the maximum displacement\(x = A\), i.e.,\(v = 0\).

The total energy of a simple harmonic oscillator can be calculated as:

\(\begin{aligned}{c}E &= \frac{1}{2}m{v^2} + \frac{1}{2}k{x^2}\\E &= \frac{1}{2} \times \left( {2.7\;{\rm{kg}}} \right) \times {\left( 0 \right)^2} + \frac{1}{2} \times \left( {310\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}} \right) \times {\left( {0.020\;{\rm{m}}} \right)^2}\\E &= 0.062\;{\rm{N}} \cdot {\rm{m}} \times \left( {\frac{{1\;{\rm{J}}}}{{{\rm{1}}\;{\rm{N}} \cdot {\rm{m}}}}} \right)\\E &= 0.062\;{\rm{J}}\end{aligned}\)

When\(v = 0\), you get the maximum deflection\({x_{\max }}\).

\(E = \frac{1}{2}kx_{\max }^2\) … (i)

After substituting the given values in equation (i), you get:

\(\begin{aligned}{c}\left( {0.062\;{\rm{J}}} \right) &= \frac{1}{2} \times \left( {310\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}} \right) \times x_{\max }^2\\\left( {0.124\;{\rm{J}}} \right) = \left( {310\;{{\rm{N}} \mathord{\left/{\vphantom {{\rm{N}} {\rm{m}}}} \right.} {\rm{m}}}} \right) \times x_{\max }^2\\x_{\max }^2 &= 4 \times {10^{ - 4}}\;{{\rm{m}}^2}\\{x_{\max }} &= 0.02\;{\rm{m}}\end{aligned}\)

Thus, the amplitude of the motion is \(0.02\;{\rm{m}}\).

(b)

When\(x = 0\), you get the maximum speed attained by the object.

\(E = \frac{1}{2}kv_{\max }^2\) … (ii)

After substituting the given values in equation (ii), you get:

\(\begin{aligned}{c}\left( {0.062\;{\rm{J}}} \right) &= \frac{1}{2} \times v_{\max }^2\\\left( {0.124\;{\rm{J}}} \right) &= v_{\max }^2\\{v_{\max }} &= 0.352\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the maximum speed attained by the object is \(0.352\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).