Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. What are the angular frequency w, the frequency, and the period of the motion?

Answer 1

Answer:

ω = 22.36 Hz

f = 3.56 Hz

T = 0.28 s.      

Explanation:

a) The angular frequency (ω), can be calculated using the following equation:

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{F}{x*m}}$  

Where:

k: is the spring constant = F/x

m: is the mass of the particle = 0.500 kg

F: is the force applied = 7.50 N      

x: is the displacement = 3.00 cm = 0.03 m            

$\omega = \sqrt{\frac{7.50 N}{0.03 m*0.500 kg}} = 22.36 s^{-1} = 22.36 Hz$    

Therefore, the angular frequency of the motion is 22.36 Hz.

b) To find the frequency (f) we can use the next equation:

$f = \frac{\omega}{2 \pi} = \frac{22.36 Hz}{2 \pi} = 3.56 Hz$

Hence, the frequency of the motion is 3.56 Hz.

c) The period (T) is equal to:

$T = \frac{1}{f} = \frac{1}{3.56 Hz} = 0.28 s$

Therefore, the period of the motion is 0.28 s.

I hope it helps you!