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For a 2.7kg mass oscillating in simple harmonic motion (spring constant: 360N/m) with an amplitude of oscillations measured at 3.4cm.

How do I calculate the total mechanical energy, maximum speed and at what positive displacement, the mass has half the maximum speed.

Penny
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  • Welcome to math stack exchange! Consider the energy's appearing in the question. Use that the sum of all energy's is constant in a closed system. – Peter Apr 29 '16 at 10:29
  • In this forum, you should show what you have tried and where you got stuck. It is easier to answer the question if we know the problems you have with the exercise. – Peter Apr 29 '16 at 10:30
  • I haven't done anything because I don't know how to proceed. If you like, you could solve a different problem the requires the same process so I can learn and solve this one myself. – Penny Apr 29 '16 at 11:17
  • The two energy's here are $\frac{D}{2}s^2$ and $\frac{m}{2}v^2$. When the energy $\frac{D}{2}s^2$ is maximal, the energy $\frac{m}{2}v^2$ is zero. So, the total energy is $0.208 Nm$. Use this to calculate the velocity at the point, where $s=0$. – Peter Apr 29 '16 at 12:40
  • What are D and s? I don't understand. My teacher has given us lots of problems like this to solve but none of us know the equations involved (with all the defined terms). I have been trying to find a good website detailing them but I can't so I thought I would ask here. – Penny Apr 29 '16 at 13:47

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For one-dimensional simple harmonic motion, the equation of motion is $$ \ddot x(t)+\omega^2 x(t)=0 $$ where $ \omega = \sqrt{\frac{k}{m}}$, $m$ is the inertial mass of the oscillating body, $x(t)$ is its displacement from the equilibrium position, and $k$ is the spring constant for a mass on a spring.

Solving the differential equation above produces a solution that is a sinusoidal function: $$ x(t) = A\cos\left(\omega t + \varphi\right), $$ where $A$ is the amplitude (maximum displacement from the equilibrium position), $ω = 2πf$ is the angular frequency, and $\varphi$ is the phase.

The velocity and acceleration as a function of time are \begin{align} v(t) &= \dot x(t) = - A\omega \sin(\omega t+\varphi)\\ a(t) &= \ddot x(t)=-\omega^2 x(t)= - A \omega^2 \cos( \omega t+\varphi). \end{align} The maximum speed is $v_{\max}= \omega A$ at equilibrium point and the maximum acceleration $a_{\max}= \omega^2 A$ at extreme points.

The velocity as function of displacement is obtained from \begin{align} v^2(t)&=\left[- A\omega \sin(\omega t+\varphi)\right]^2\\& =\omega^2 A^2\sin^2(\omega t+\varphi)=\omega^2 A^2\left[1-\cos^2(\omega t+\varphi)\right]\\ &=\omega^2 (A^2-x^2(t)) \end{align} that is $$ v(x)=\omega\sqrt{A^2-x^2} $$ So the positive displacement $\tilde x$ at which the mass has half the maximum speed will be found solving $$ v\left(\tilde x\right)=\omega\sqrt{A^2-\tilde x^2}=\frac{1}{2}v_{\max}=\frac{1}{2} \omega A $$ that is $$\omega^2 (A^2-\tilde x^2)= \frac{\omega^2 A^2}{4}\quad \Longrightarrow \tilde x=\frac{\sqrt 3}{2}A$$

The kinetic energy $K$ of the system at time $t$ is $$ K(t) = \frac{1}{2} mv^2(t) = \frac{1}{2}m\omega^2A^2\sin^2(\omega t + \varphi) = \frac{1}{2}kA^2 \sin^2(\omega t + \varphi), $$ and the potential energy is $$ U(t) = \frac{1}{2} k x^2(t) = \frac{1}{2} k A^2 \cos^2(\omega t + \varphi). $$ The total mechanical energy has a constant value $$ E = K + U = \frac{1}{2} k A^2. $$

alexjo
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