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?
a) 5.23 cm
b) 16.6 cm
c) 5.41 cm
d) 8.84 cm
e) 11.6 cm

Answer 1

Answer:

Option b: 16.6 cm.

Explanation:

The position of the mass at 2.50 s can be found using the simple harmonic motion equation:

$x_{t} = Acos(\omega t)$

Where:

A: is the amplitude = 17.5 cm

ω: is the angular frequency = 2πf

t = 2.50 s

$x_{t} = Acos(\omega t) = 17.5cos(2\pi*8.38*2.50) = 16.6 cm$

Therefore, the correct answer is option b: 16.6 cm.

I hope it helps you!