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

`t_1=(5\pi)/(17) -(5\pi)/(34) \approx 0.462 s\\\\t_2=(10\pi)/(17) -(5\pi)/(34) \approx1.39s`

Step-by-step 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-(\pi)/(2) ,\hspace{7}n\in Z`

Now:

Rewrite 3.4 as:

`3.4=(17)/(5)`

So:

`(17)/(5) t= \pi n -(\pi)/(2) \\\\Hence\\\\t=(5\pi n)/(17) -(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=(5\pi n)/(17) -(5\pi)/(34),\hspace{7}n\in Z\\`

The first time is when n=1, so:

`t_1=(5\pi(1))/(17) -(5\pi)/(34)=(5\pi)/(17) -(5\pi)/(34) \approx 0.462 s`

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

`t_2=(5\pi(2))/(17) -(5\pi)/(34)=(10\pi)/(17) -(5\pi)/(34) \approx1.39s`