Question

A 0.26 kg mass at the end of a spring oscillates 2.0 times per second with an amplitude of 0.13 m.

A) Determine the speed when it is 0.12 m from equilibrium.

B)Determine the total energy of the system.

C) Determine the equation describing the motion of the mass, assuming that at t=0, x was a maximum, pick the correct answer below.

- x(t) = (0.13m)cos[(2.0Hz)t]

- x(t)=(0.13m)cos[2π(2.0Hz)t]

- x(t)=(0.13m)sin[2π(2.0Hz)t]

- x(t)=(0.065m)cos[2π(2.0Hz)t]

Answer 1

here,

mass , m = 0.26 kg

frequency , f = 2 Hz

amplitude , A = 0.13 m

a)

let the spring constant be K

f = 1/2*pi * sqrt(K/m)

2 = 1/2pi * sqrt(K/0.26)

K = 41.02 N/m

at x1 = 0.12 m let the speed be v1

using conservation of energy

0.5 * K * A^2 = 0.5 * K * x1^2 + 0.5 * m * v1^2

41.02 * 0.13^2 = 41.02 * 0.12^2 + 0.26 * v^2

v = 0.13 m/s

the speed of mass is 0.13 m/s

b)

the total energy of the system , TE = 0.5 * K * A^2

TE = 0.5 * 41.02 * 0.13^2 J = 0.35 J

c)

as at t = 0 , x was a maximum

x(t) = A * cos(2*pi*f*t)

x(t) = (0.13 m) * cos(2*pi*(2Hz)t)