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 6.10 N is applied. A 0.420-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? ? = 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)By hooke's law
F=KX =>K=F/X
K=8.3/0.03 =276.67 N/m
b) =(K/m)

angular frequency

=(276.67/0.52) =23.07 rad/s

period T=2/

T=2/23.07 =0.272 s

frequency f=1/T=3.67Hz

c)Total energy of the system

E=(1/2)Kxo2 =(1/2)*276.67*0.052 =0.346J

d)At maximum displacement from equilibrium, x = A = Amplitude. Plus, it is at rest (for and instant) implying all the energy is elastic potential energy.

E=(1/2)KA2

A=(2E/K)

A=(2*0.346/276.67) =0.05m or 5cm

e)Vmax is at equilibrium (x = 0) where all the energy is kinetic

E=(1/2)mVmax2 =>Vmax=(2E/m)

Vmax =(2*0.346/0.52) =1.15m/s2

the maximum acceleration of the particle occurs when the spring exerts maximum forces on it (when spring

is stretched or compressed to its maximum). |x|=A

amax =KA/m =276.67*0.05/0.52 =26.6m/s2

f) the displacement at any time is:

X=Acost =Acos[(K/m)t]

at t=0.5s

X=0.05cos(0.5*23.07)=0.0257m or 2.57cm

g)X=Acost

V=dX/dt =-Asint

at t=0.5

V=-(0.05)(23.07)sin(23.07*0.5)=0.9897m/s0.99m/s

a=dV/dt=-A2cost

at t=0.5s

a=-0.05*(23.07)2cos(23.07*0.5)

a=-13.67m/s2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.