\text{Chapter 15}

\text{A block of mass $2.0 \mathrm{~kg}$ attached to a spring oscillates in simple harmonic motion along the}

\text{$x$ axis. The limits of its motion are $x=-20 \mathrm{~cm}$ and $x=+$ $20 \mathrm{~cm}$ and it goes from one of}

\text{ these extremes to the other in $0.25 \mathrm{~s}$. The mechanical energy of the block-spring system is:}

\text{A $2.0-\mathrm{kg}$ mass connected to a spring of force constant $8.0 \mathrm{~N} / \mathrm{m}$ is displaced $5.0 \mathrm{~cm}$ from}

\text{ its equilibrium position and released. It oscillates on a horizontal, frictionless surface.}

\text{ Find the speed of the mass when it is at $3.0 \mathrm{~cm}$ from its equilibrium position.}

\text{The displacement of a particle moving in simple harmonic motion is given by the equation:}

\text{ $x(t)=2.0 \cos (6.0 t)$, where $x$ is in meters and $t$ is in seconds. }

\text{What is the maximum acceleration of the particle and where does it occur?}

\text{The displacement of a particle moving in simple harmonic motion is given by the equation:}

\text{ $x(t)=2.0 \cos (6.0 t)$, where $x$ is in meters and $t$ is in seconds. }

\text{What is the maximum acceleration of the particle and where does it occur?}

x(t)=2 \cos(6.0 t)

\omega=6 \text{ rad/s}

a(t)=-\omega^2 x(t)

a(t) \text{ will be maximum for } x(t)=-x_m

a(t)=\omega^2 xm=6^2\times 2=72 \text{ $m/s^2$}

\text{ for $x(t)$ minimum}