An object with mass 2.7 kg is executing simple harmonic motion, attached to a spring with spring constant k =310 N/m.When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s. (a) Calculate the amplitude of the motion. (b) Calculate the maximum speed attained by the object.

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 \(0.352\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).