A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.70 N is applied. A 0.460-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.)

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

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

w = ________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

Explanation / Answer

a) F=kx.
7.70=k(3*10^-2)
k=256.67 Nm^-1

b) W^2=k/m
W^2=256.67/0.460
W=23.62 rad/s

c) W=2(pi)/T
T=W/2(pi)
T=23.62/2(pi)
T=3.75 s.
Frequency=1/period
=1/3.75
=0.0.266 Hz

d)A=5.00 cm.

e)V=-AWsinWt

max velocity is when t=1.875 s
A=5*10^-2 m
W=23.62
V=-(5*10^-2)(23.62)sin(23.62 )
V=-0.4732 ms^-1

a=-AW^2cosWt
max acceleration is when t =0
a=-AW^2cosWt
a=-(5*10^-2)(23.62)^2cos(23.62)
a=-25.558ms^-2

f)x=AcosWt
x=(5*10^-2)(cos(23.62))(0.500)
x=0.04894 m

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