Question

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. the mass is pulled away from the equilibrium position (x = 0) a distance of 17.5 cm and released. it then oscillates in simple harmonic motion with a frequency of 8.38 hz. at what position, measured from the equilibrium position, is the mass 2.50 seconds after it is released?

Answer 1

For a simple harmonic motion, the position of the mass at any time t is given by
$x(t)=A cos (\omega t)$
where 
A is the amplitude of the motion (in this problem, A=17.5 cm)
$\omega$ is the angular frequency of the oscillator
t is the time

The angular frequency of the motion in the problem is given by
$\omega = 2 \pi f= 2 \pi (8.38 Hz) = 52.7 rad/s$

And so, we can find the position x of the mass (with respect to the equilibrium position) at time t=2.50 s:
$x(2.50 s)=(17.5 cm) \cos ( (52.7 rad/s)(2.50 s))=17.2 cm$

Answer 2

16.64cm is the correct answer.

Don't round ω when entering cos(ω*T) in your calculator