Consider a mass M = 4.9 kg hanging from a spring with force constant k = 6900 N/

Question

Consider a mass M = 4.9 kg hanging from a spring with force constant k = 6900 N/m.Let x denote the vertical coordinate of the mass; the mass is inequilibrium at x = 0; positive direction is up. At time t0 = 0 the mass is moved to x0 = +5 cm and given initial speed v0 = 1.4 m/s.(Mind the units!) Subsequently, it oscillates
according to x(t) = A × sin(t + 0).
1.What is the amplitude A of this oscillation?
Answer in units of cm.

2.What is the initial phase 0? Give an
answer between and +.
Answer in units of rad.

3.What is the total energy of the oscillation?
Answer in units of J.

4.When is the kinetic energy of the oscillator
is equal to twice the potential energy — K =
2U — for the first time (after T0 = 0)?
Answer in units of s.

Explanation / Answer

1. w = sqrt(k/m) = 37.5 rad/s

v^2 = w^2 (A^2 - x^2)

1.4^2 = 37.5^2(A^2 - 0.05^2)

A = 0.062 m Or 6.2 cm

2. x = A sin(wt + phi)

5 = 6.2 sin(phi)

phi = 2.20 rad

3. total energy = kA^2 /2 = 6900 x 0.062^2 / 2

= 13.3 J

4. 4.9 v^2 / 2 = 6900 x^2

v^2 = 2816.33 x^2

v^2 = w^2 ( A^2 - x^2)

2816.33 x^2 / 37.5^2 = A^2 - x^2

x = 0.577 A

0.577 = sin(37,5 t + 2.20)

t = 8.70 x 10^-3 s

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