Question

Picture the Problem The position of the particle as a function of time is given byx - Acos(or 5). We're given the amplitude A of the motion and can use the initial position of the particle to determine the phase constant & Once we've determined these quantities, we can express the distance traveled Δχ during any interval of time. Express the position of the particle asx-(12 cm)cos(or +8) a function of t Find the angular frequency of the particle's motion T8.0s 4 Relate the initial position of the particle to the amplitude and phase constant: Solve for -1 A 2 Substitute in equation (1) to obtain: π-1·, π 4 ) Express the distance the particle travels in terms off and 12 cm)cos

Why did the amplitude disappear in the highlighted part?

Answer 1

As per the equation, the distance particle travels is,

$\Delta x = \left | (12 cm) cos \left( \frac{\pi}{4}t_f + \frac{\pi}{2} \right ) - (12 cm) cos \left( \frac{\pi}{4}t_i + \frac{\pi}{2} \right )\right |$

Since the 12 cm term is common, taking it outside we get,

$\Rightarrow \Delta x = \left | (12 cm) \left [ cos \left( \frac{\pi}{4}t_f + \frac{\pi}{2} \right ) - cos \left( \frac{\pi}{4}t_i + \frac{\pi}{2} \right ) \right ] \right |$

So the amplitude term has been taken outside in the next step, which is why it is missing in that second term.