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 6.50 N is applied. A 0.410-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 , 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.)

PLEASE ANSWER ALL PARTS THANK YOU

Explanation / Answer

a)

force constant

K=F/x=6.5/0.03=216.67 N/m

b)

angular frequency

W=sqrt(K/m)=sqrt(216.67/0.41)=23 rad/sec

frequency

f=W/2pi=23/2pi=3.66 Hz

time period

T=1/f=1/3.66=0.2733 s

c)

total energy

E=(1/2)KA2=(1/2)*216.67*0.052=0.271 J

d)

amplitude

A=5 cm

e)

maximum velocity

Vmax=WA=23*0.05=1.15 m/s

maximum acceleration

amax=W2A=232*0.05=26.42 m/s2

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