\text{Chapter 15}

\begin{align*} &\text{A 2.0-kg mass connected to a spring of force constant } 8.0~\mathrm{N/m} \text{ is } \text{displaced } 5.0~\mathrm{cm} \text{from its}\\ & \text{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} \text{ from its equilibrium position.} \\[4pt] &\\ &\text{A) } 0.20~\mathrm{m/s} \\ &\text{B) } 0.04~\mathrm{m/s} \\ &\text{C) } 0.12~\mathrm{m/s} \\ &\text{D) } 0.08~\mathrm{m/s} \\ &\text{E) } 0.32~\mathrm{m/s} \end{align*}

\begin{align*} &\text{A 2.0-kg mass connected to a spring of force constant } 8.0~\mathrm{N/m} \text{ is } \text{displaced } 5.0~\mathrm{cm} \text{ from its}\\ & \text{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} \text{ from its equilibrium position.} \\[4pt] \end{align*}

\text{The total energy is: }

E=\frac{1}{2}k x_m^2

E=\frac{1}{2}k x_m^2 =\frac{1}{2}k x^2+\frac{1}{2} m v^2

v=\sqrt{\frac{k(x_m^2-x^2)}{m}}

\Rightarrow

k=8.0 \mathrm{~N/m}

x_m=5.0\mathrm{~cm}

v=0.08 \mathrm{~m/s}

m=2.0 \mathrm{~kg}

\begin{align*} &\text{The displacement of a particle moving in simple harmonic motion is given by the equation} \\ & x(t)=2.0 \cos(6.0 t), \text{ where } x \text{ is in meters and } t \text{ is in seconds.} \\ &\text{What is the maximum acceleration of the particle and where does it occur?} \\[4pt] &\\ &\text{A) } 72~\mathrm{m/s^2} \text{ at maximum negative displacement} \\ &\text{B) } 72~\mathrm{m/s^2} \text{ at maximum positive displacement} \\ &\text{C) } 12~\mathrm{m/s^2} \text{ at maximum negative displacement} \\ &\text{D) } 12~\mathrm{m/s^2} \text{ at maximum positive displacement} \\ &\text{E) } 2.0~\mathrm{m/s^2} \text{ at zero displacement} \end{align*}

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

\omega=6 \text{ rad/s}

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

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

\begin{align*} &\text{The displacement of a particle moving in simple harmonic motion is given by the equation} \\ & x(t)=2.0 \cos(6.0 t), \text{ where } x \text{ is in meters and } t \text{ is in seconds.} \\ &\text{What is the maximum acceleration of the particle and where does it occur?} \\[4pt] \end{align*}

\text{The acceleration is:}

\begin{align*} &\text{A block of mass } 2.0~\mathrm{kg} \text{ attached to a spring oscillates in simple } \text{harmonic motion along the } \\ & x\text{-axis. The limits of its motion are } x=-20~\mathrm{cm} \text{ and } x=+20~\mathrm{cm} \text{ and it goes from one} \\ &\text{extreme to the other in } 0.25~\mathrm{s}. \text{ The mechanical energy of the} \text{block-spring system is:} \\[4pt] &\text{A) } 6.3~\mathrm{J} \\ &\text{B) } 1.2~\mathrm{J} \\ &\text{C) } 2.5~\mathrm{J} \\ &\text{D) } 5.3~\mathrm{J} \\ &\text{E) } 4.1~\mathrm{J} \end{align*}

\begin{align*} &\text{A block of mass } 2.0~\mathrm{kg} \text{ attached to a spring oscillates in simple } \text{harmonic motion along the } \\ & x\text{-axis. The limits of its motion are } x=-20~\mathrm{cm} \text{ and } x=+20~\mathrm{cm} \text{ and it goes from one} \\ &\text{extreme to the other in } 0.25~\mathrm{s}. \text{ The mechanical energy of the } \text{block-spring system is:} \\[4pt] \end{align*}

\text{From one extreme to the other, we only take half the period: } T/2=0.25 \mathrm{~s}

T=0.5 \mathrm{~s}

\text{The total energy is: }

E=\frac{1}{2}k x_m^2

\text{For a harmonic oscillator, we have:}

E=2 \pi^2 \frac{m x_m^2}{T^2}=2\pi^2 \frac{2.0\times 0.2^2}{0.5^2}=6.3 \mathrm{~J}

T=2 \pi \sqrt{\frac{m}{k}}

k=4\pi^2 \frac{m}{T^2}

\cdots \textcircled{1}

\cdots \textcircled{2}

\text{replacing } \textcircled{2} \text{ in }\textcircled{1},

\text{(Answer A)}

\begin{align*} &\text{A block attached to a spring undergoes a simple harmonic motion on a horizontal frictionless } \\ &\text{surface. Its mechanical energy is } 40~\mathrm{J}. \\ &\text{When the displacement is half the amplitude, the kinetic energy is:} \\[4pt] &\\ &\text{A) } 15~\mathrm{J} \\ &\text{B) zero} \\ &\text{C) } 30~\mathrm{J} \\ &\text{D) } 25~\mathrm{J} \\ &\text{E) } 40~\mathrm{J} \end{align*}

\begin{align*} &\text{A block attached to a spring undergoes a simple harmonic motion on a horizontal frictionless } \\ &\text{surface. Its mechanical energy is } 40~\mathrm{J}. \\ &\text{When the displacement is half the amplitude, the kinetic energy is:} \\[4pt] \end{align*}

E=\frac{1}{2}k x_m^2 \text{ and also }E=\frac{1}{2}k x^2+\frac{1}{2} m v^2

\text{For }x=\frac{x_m}{2}

E=\frac{1}{2}k \frac{x_m^2}{4}+\underbrace{\frac{1}{2}m v^2}_{K}

E=\frac{E}{4}+K

\Rightarrow

K=\frac{3E}{4}=30\mathrm{~J}

\Rightarrow

\begin{align*} &\text{A block-spring system oscillates with simple harmonic motion according to the equation } \\ & x = 0.20 \cos(10t + \pi/2),\text{ where } x \text{ is in m and } t \text{ is in s. The mass of the block is } 2.0~\mathrm{kg}.\\ &\text{Find the total} \text{energy of the system.} \\ &\\ &\text{A) } 4.0~\mathrm{J} \\ &\text{B) } 100~\mathrm{J} \\ &\text{C) } 8.0~\mathrm{J} \\ &\text{D) } 10~\mathrm{J} \\ &\text{E) } 15~\mathrm{J} \end{align*}

\begin{align*} &\text{A block-spring system oscillates with simple harmonic motion according to the equation } \\ & x = 0.20 \cos(10t + \pi/2),\text{ where } x \text{ is in m and } t \text{ is in s. The mass of the block is } 2.0~\mathrm{kg}.\\ &\text{Find the total} \text{ energy of the system.} \\ \end{align*}

x = 0.20 \cos(10t + \pi/2)

\omega=10\mathrm{~\frac{rad}{s}}

x_m=0.20 \mathrm{~m}

m=2.0 \mathrm{~kg}

\omega=\sqrt{\frac{k}{m}}\Rightarrow k=m\omega^2

E=\frac{1}{2}k x_m^2=\frac{1}{2} m\omega^2 x_m^2=0.5\times 2\times 10^2\times 0.2^2= 4\mathrm{~J}

\text{Another method}

\text{since the total energy is constant, we express it the origin, where $v$ is maximum }

E=\underbrace{\frac{1}{2}k x^2}_{=0}+\frac{1}{2}m v_m^2

\text{We have }v_m=\omega x_m

E=\frac{1}{2}m \omega x_m^2

\text{(which is the same expression we got previously)}

\text{(Answer A)}

\begin{align*} &\text{A physical pendulum consists of a uniform solid disk (radius } R=10.0~\mathrm{cm}) \text{ supported in a} \\ &\text{vertical plane by a pivot located at a distance } d=5.0~\mathrm{cm}\text{from the center of the disk. }\\ &\text{The disk is made to oscillate in a simple} \text{harmonic motion of period } T.\text{ Find } T. \\[4pt] &\\ &\text{A) } 1.8~\mathrm{s} \\ &\text{B) } 1.4~\mathrm{s} \\ &\text{C) } 1.0~\mathrm{s} \\ &\text{D) } 0.38~\mathrm{s} \\ &\text{E) } 0.78~\mathrm{s} \end{align*}

\begin{align*} &\text{A physical pendulum consists of a uniform solid disk (radius } R=10.0~\mathrm{cm}) \text{ supported in a} \\ &\text{vertical plane by a pivot located at a distance } d=5.0~\mathrm{cm}\text{ from the center of the disk. }\\ &\text{The disk is made to oscillate in a simple } \text{harmonic motion of period } T.\text{ Find } T. \\[4pt] \end{align*}

\text{x}

\text{we have }d=\frac{R}{2}

\text{The disk is pivoting about a point away from its center of mass}

T=2 \pi \sqrt{\frac{I}{m g h}}

I=I_0+mh^2=\frac{1}{2}m R^2+m\frac{R^2}{4}=3m\frac{R^2}{4}

\text{(Axis parallel theorem)}

h=\frac{R}{2}

T=2 \pi \sqrt{\frac{I}{m g h}}=2 \pi \sqrt{\frac{3/4 mR^2}{m g R/2}}=2 \pi \sqrt{\frac{3R}{2g}}=2\pi\sqrt{\frac{3\times0.1}{2\times 9.8}}=0.78 \mathrm{~s}

\text{(Answer E)}

\begin{align*} &\text{A simple pendulum of length } 1.55~\mathrm{m} \text{ has a period } (T) \text{ on the surface of Earth. What is the} \\ &\text{length of the pendulum to have the same period } (T) \text{on the surface of the Moon where } \\ & g = 1.67~\mathrm{m/s^2}\text{?} \\[4pt] &\\ &\text{A) } 0.53~\mathrm{m} \\ &\text{B) } 2.64~\mathrm{m} \\ &\text{C) } 0.26~\mathrm{m} \\ &\text{D) } 1.32~\mathrm{m} \\ &\text{E) } 5.28~\mathrm{m} \end{align*}

\begin{align*} &\text{A simple pendulum of length } 1.55~\mathrm{m} \text{ has a period } (T) \text{ on the surface of Earth. What is the} \\ &\text{length of the pendulum to have the same period } (T) \text{on the surface of the Moon where } \\ & g = 1.67~\mathrm{m/s^2}\text{?} \\[4pt] \end{align*}

T'=2 \pi \sqrt{\frac{L}{g'}}

T=2 \pi \sqrt{\frac{L}{g}}

T'=2 \pi \sqrt{\frac{L'}{g'}}

\text{On earth}

\text{On the Moon}

T=T' \Rightarrow \frac{L}{g}=\frac{L'}{g'}

\Rightarrow {L'}= L\frac{g'}{g}=1.55\times \frac{1.67}{9.8}=0.26 \mathrm{~m}

\text{(Answer C)}