Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.10 N is applied. A 0.440-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. (Assume that the direction of the initial displacement is positive. Use the exact values you enter to make later calculations.)

(a) What is the force constant of the spring?
N/m

(b) What are the angular frequency ω, the frequency, and the period of the motion?

ω =	rad/s
f =	Hz
T =	s

(c) What is the total energy of the system?
J

(d) What is the amplitude of the motion?
cm

(e) What are the maximum velocity and the maximum acceleration of the particle?

vmax =	m/s
amax =	m/s2

(f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s.
cm

(g) Determine the velocity and acceleration of the particle when t = 0.500 s. (Indicate the direction with the sign of your answer.)

v =	m/s
a =	m/s2

Answer 1

a)

using hooke's law

k = F/x = 7.1 / 0.03 = 236.67 N/m

b)

w^2 = k/m = 236.67 / 0.44

w = 23.192 rad/s

frequency

f = w / (2 pi) = 3.691 Hz

time period

T = 1/f = 0.271 s

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c)

TE = 0.5 kA^2 = 0.5* 236.67*0.5^2 = 0.2958 J

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d)

A = 5 cm

=====

e)

Vmax = A w = 0.05* 23.192 = 1.16 m/s

amax = A w^2 = 26.893 m/s^2

======

f)

x = A cos ( wt)

x = 5 cos ( 23.192* 0.5) = 2.285 cm

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g)

v = dx/dt = - Aw sin (wt) = - 0.05* 23.192 sin ( 23.192* 0.5) = +0.957 m/s

a = dv/dt = - 0.05^2 *23.192 cos ( 23.192* 0.5) = - 15.195 m/s^2