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 7.00 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?

(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

A)
force F = K(0.03)
K = 7/0.03 = 233.33 N/m
mass m = 0.420 kg
------------------------
B)
Amplitude A = 0.05 m
ang freq w = sqrt(k/m) =23.57 rad/s
freq f = w/2pi = 3.753 hz
peroi T = 1/f = 0.266 sec
--------------------------
C)
total energy =0.5*KA^2 = 0.2916 J
--------------------------
D)
amplitude A =0.05 m
---------------
e)
Vmax = WA = 23.57*0.05 =1.1785
a = -W^2A = - 27.7 m/s^2
---------------------------------
f)

x = 0.05cos (23.57t)
at t =0.5 sec
x =0.03549 m

-----------------------------------------

g)
velcoity
v = -wAsin(wt) = 0.833
a = -w^2 A coswt =-19.72 m/s^2

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