A \(0.650\;{\rm{kg}}\) mass oscillates according to the equation \(x = 0.25\sin \left( {4.70\;{\rm{t}}} \right)\) where \({\rm{x}}\) is in meters and \({\rm{t}}\) is in seconds. Determine (a) the amplitude, (b) the frequency, (c) the period, (d) the total energy, and (e) the kinetic energy and potential energy when \({\rm{x}}\)is15 cm

In this problem, the kinetic energy and potential energy are determined by applying the principle of conservation of total energy.

The displacement equation for SHM is\(x = 0.25\sin \left( {4.70t} \right)\).

The time is\(t\).

The mass is\({\rm{m}} = 0.650\;{\rm{kg}}\).

The amplitude of the oscillation will be equivalent to the maximum displacement. So, the amplitude is calculated as:

\(\begin{aligned}{c}A &= {x_{{\rm{max}}}}\\ &= 0.25\;{\rm{m}}\end{aligned}\)

Hence, amplitude of the mass is\(0.25\;{\rm{m}}\).

The frequency of oscillation is calculated as:

\(f = \frac{\omega }{{2{\rm{\pi }}}}\)

Substitute the values in the above relation.

\(\begin{aligned}{c}f = \frac{{4.70\;{{\rm{s}}^{ - 1}}}}{{2{\rm{\pi }}}}\\f = 0.748\;{\rm{Hz}}\end{aligned}\)

Hence, the frequency of oscillation is 0.748 Hz

The period is calculated as:

\(T = \frac{1}{f}\)

Substitute the values in the above relation.

\(\begin{aligned}{c}T = \frac{1}{{0.748\;{\rm{Hz}}}}\\T = 1.34\;{\rm{s}}\end{aligned}\)

Hence, the period of oscillation is 1.34 s.

The total energy is calculated as:

\({{\rm{E}}_{\rm{T}}} = \frac{1}{2}{\rm{mv}}_{{\rm{max}}}^{\rm{2}}\)

Substitute the values in the above relation.

\begin{aligned}{c}{E_{\rm{T}}} &= \frac{1}{2}m{\left( {\omega A} \right)^2}\\{E_{\rm{T}}} &= \frac{1}{2}\left( {0.65\;{\rm{kg}}} \right){\left( {4.70\;{{\rm{s}}^{ - 1}}} \right)^2}{\left( {0.25\;{\rm{m}}} \right)^2}\\{E_{\rm{T}}} &= 0.45\;{\rm{J}}\end{aligned}

Hence, total energy of the mass is \(0.45\;{\rm{J}}\).

According to the conservation of energy, total energy remains constant.

The potential energy is calculated as:

\begin{aligned}{c}PE &= \frac{1}{2}k{x^2}\\ &= \frac{1}{2}m{\left( {\omega x} \right)^2}\\ &= \frac{1}{2}\left( {0.65\;{\rm{kg}}} \right){\left( {4.70\;{{\rm{s}}^{ - 1}}} \right)^2}{\left( {0.15\;{\rm{m}}} \right)^2}\\ &= 0.16\;{\rm{J}}\end{aligned}

The kinetic energy is calculated as:

\begin{aligned}KE &= {E_{\rm{T}}} - PE\\ &= 0.45\;{\rm{J}} - {\rm{0}}{\rm{.16}}\;{\rm{J}}\\ &= 0.29\;{\rm{J}}\end{aligned}

Hence, potential and kinetic energy of the mass is \(0.16\;{\rm{J}}\) and \(0.29\;{\rm{J}}\), respectively.