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 8.30 N is applied. A 0.510-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 (?), the frequency, and the period of the motion?

(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?

(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.)

Explanation / Answer

given values are

F=8.30N

x=3cm=0.03m

m=0.510-kg

x = 5.00 cm at t=0 ie--0.05m

a) F=kx

k=F/x

=8.30/0.03

k=276.6N/m

b)T = 2??(m/k)

=2*3.14*undrt(0.510/276.6)

T=0.269 sec

f = 1/T

=1/0.269

f=3.71Hz

? =?(k/m)

=undrt(276.6/0.510)

=23.28rad/s

c)E=(kx^2)/2

=(276.6(0.05)^2)/2

E=0.345J

d)Obvious. When you stretch it and let it go, that's the amplitude.

so amplitute=0.05m

e) v = ?A

=23.28*0.05

v=1.164 m/sec

a = (kA)/m

=(276.6*0.05)/0.510

a=27.11m/s^2

f)x = -Acos?t

=-0.05cos(23.28*0.5)

x=-0.03m

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