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 6.70 N is applied. A 0.440-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.) What is the force constant of the spring? ______ N/m What are the angular frequency omega, the frequency, and the period of the motion? omega = ____ rad/s f = _____ Hz T = ______ s What is the total energy of the system? _______ J What is the amplitude of the motion? __________ cm What are the maximum velocity and the maximum acceleration of the particle? v_max = _____________ m/s a_max = _____________ m/s^2 Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. ________________ cm 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/s^2

Explanation / Answer

a)We know that

F = -kx

k = F/x = 6.7/0.03 = 223.33 N/m

Hence, k = 233.33 N/m

b)we know that,

w = sqrt(k/m) = sqrt (223.33/0.44) = 22.53 rad/s

Hence, w = 22.53 rad/s

2 pi f = w => f = w/ 2 pi = 22.53/2 x 3.14 = 3.59 Hz

Hence, f = 3.59 Hz

We know that, f = 1/T=> T = 1/f

T = 1/3.59 = 0.28 s

Hence, T = 0.28 s

c)E = 1/2 k x^2

E = 0.5 x 223.33 x 0.03^2 = 0.1 J

E = 0.1 J

d)v = x sqrt(k/m)

v = 0.03 x sqrt(223.33/0.44) = 0.68 m/s

Hence, v = 0.68 m/s

a = v/t = 0.68/0.28 = 2.43 m/s^2

Hence, a = 2.43 m/s^2

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