Question

For a particle undergoing simple harmonic motion described by x(t) = (4.5 cm) cos ((3.4 Hz) t)x(t)=(4.5cm)cos((3.4Hz)t) Find the first two times that the particle crosses x(t) = 0.x(t)=0.

Answer 1

Answer:

$t_1=\frac{5\pi}{17} -\frac{5\pi}{34} \approx 0.462 s\\\\t_2=\frac{10\pi}{17} -\frac{5\pi}{34} \approx1.39s$

Explanation:

No matter the coeficient of the cosine function, the function will always be zero as long as the following is true:

$cos(t)=0\\\\for\\\\t=\pi n-\frac{\pi}{2} ,\hspace{7}n\in Z$

Now:

Rewrite 3.4 as:

$3.4=\frac{17}{5}$

So:

$\frac{17}{5} t= \pi n -\frac{\pi}{2} \\\\Hence\\\\t=\frac{5\pi n}{17} -\frac{5\pi}{34},\hspace{7}n\in Z$

Therefore the particle crosses the x-axis (x(t)=0) :

$x(t)=4.5cos((3.4)t)=0,\hspace{10}When\\\\t=\frac{5\pi n}{17} -\frac{5\pi}{34},\hspace{7}n\in Z$

The first time is when n=1, so:

$t_1=\frac{5\pi(1)}{17} -\frac{5\pi}{34}=\frac{5\pi}{17} -\frac{5\pi}{34} \approx 0.462 s$

And the second time is when n=2, so:

$t_2=\frac{5\pi(2)}{17} -\frac{5\pi}{34}=\frac{10\pi}{17} -\frac{5\pi}{34} \approx1.39s$