Chapter 14 - Oscillations - Exercises and Problems - Page 402: 14

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We are given the position function of a 50-gram object which is given by $$\boxed{x_{(t)}=0.02\cos\left(10t-\dfrac{\pi}{4}\right )}$$ And we know that the position of a mass undergoes a simple harmonic motion is given by $$ x_{(t)}=A\cos\left(\omega t+\phi_0\right ) $$ --- a) The amplitude is given from the formula above which is $$A=\color{red}{\bf 2.0}\;\rm cm$$ _________________________________________________ b) The period is given by $$\omega=2\pi f=\dfrac{2\pi }{T}$$ Hence, $$T=\dfrac{2\pi }{\omega}$$ Plugging from the boxed formula above; $$T=\dfrac{2\pi }{10}=\color{red}{\bf 0.63}\;\rm s$$ _________________________________________________ c) The spring constant is given by $$\omega=\sqrt{\dfrac{k}{m}} $$ So that $$k=m\omega^2 =(50\times 10^{-3})(10)^2$$ $$k=\color{red}{\bf 5.0}\;\rm N/m$$ _________________________________________________ d) The phase constant is given from the boxed formula above. $$\phi_0=\color{red}{\bf \frac{-\pi}{4}}\;\rm rad$$ _________________________________________________ e) The initial conditions which are the initial position and the initial velocity. So, at $t=0$ s, $$x_{(0)}=0.02\cos\left(10(0)t-\dfrac{\pi}{4}\right )=0.02\cos\frac{-\pi}{4}=\bf 0.01414\;\rm m$$ $$x_{(0)}=\color{red}{\bf 1.414}\;\rm cm$$ So the velocity is given by $$v_{x}=\dfrac{dx}{dt}=\dfrac{d }{dt}\left[ 0.02\cos\left(10t-\dfrac{\pi}{4}\right ) \right]$$ $$v_{x}= -0.2\sin\left(10t-\dfrac{\pi}{4}\right ) $$ So, at $t=0$ s, $$v_{0x}= -0.2\sin\left(10(0)-\dfrac{\pi}{4}\right )=\bf 0.1414\;\rm m/s $$ $$v_{0x}=\color{red}{\bf 14.14}\;\rm cm/s$$ _________________________________________________ f) The maximum speed is given by $$v_{\rm max}=A\omega=(2)(10)=\color{red}{\bf 20}\;\rm cm/s$$ _________________________________________________ g) The total energy is given by $$E_{tot}=\frac{1}{2}mv_{\rm max}^2$$ $$E_{tot}=\frac{1}{2}(50\times 10^{-3})(20\times 10^{-2})^2= \color{red}{\bf 10^{-3}}\;\rm J$$ _________________________________________________ h) The velocity at $t=0.4$ s is given by $$v_{x}= -0.2\sin\left(10t-\dfrac{\pi}{4}\right ) = -0.2\sin\left(10(0.4)-\dfrac{\pi}{4}\right )=\bf 0.01458\;\rm m/s$$ $$v_{0x}=\color{red}{\bf 1.46 }\;\rm cm/s$$