Sunday 28 February 2016

homework and exercises - Which trigonometric ratio should be used to describe simple harmonic motion as a function of time?



An object is undergoing simple harmonic motion with period 1.2s and amplitude 0.6m. At $t=0$, the object is at $x=0$. How far is the object from the equilibrium position when $t=0.480$s?




I used the displacement equation :


$$x(t)=A\cos(\omega t+\phi)$$


and also found out what the angular frequency $\omega$ is (5.2rad/s). Then I found that $\phi$ is 0. I plugged my results in the equation and when I looked at the solution they used the following equation:


$$x(t)=A\sin(\omega t)$$


How was I supposed to know to use this equation instead of the cosine equation that is written on my equation sheet.



Answer



The initial phase $\phi$ should be $-\pi/2$, or in other words, the solution should be $A\sin(\omega t)$ because $\cos(\theta-\pi/2)=\sin(\theta)$. How did you find that $\phi$ is zero? You're supposed to use the fact that $x$ is zero when $t$ is zero. Which is what let's you set $\phi $ to $-\pi/2$ (modulo $\pi$).


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