Question

In: Physics

A 0.47 kg mass at the end of a spring vibrates 6.0 times per second with...

A 0.47 kg mass at the end of a spring vibrates 6.0 times per second with an amplitude of 0.12 m. (a) Determine the velocity when it passes the equilibrium point. m/s (b) Determine the velocity when it is 0.08 m from equilibrium. m/s (c) Determine the total energy of the system. J (d) If the amplitude of oscillation were increased by a factor of 3.3, by what factor does the total energy change?

Solutions

Expert Solution

spring vibrates 6.0 times per second ---> f = 6Hz, T=(1/6)s
angular frequency is
ω = 2πf = 2π * 6 = 12π s^-1
Spring constant is
k = mω^2 = 0.47 * (12π)^2 = 667.29 N/m

(a) velocity when it passes the equilibrium point
If mass is released from extreme position (when displacement=amplitude) velocity as function of time is
v(t) = -Aω sin(ωt)
Mass passes through equilibrium point for the first time after one quarter of period, t=T/4=(1/24) s
vmax = v(t=T/4) = -0.12 * 12π sin(12π/24)
vmax = 4.52 m/s

(b) Determine the velocity when it is 0.08 m from equilibrium
In equilibrium position the mass posesses only kinetic energy so total energy of the system is:
E = Ekmax = m vmax^2 / 2
E = 0.47 * 4.52^2 / 2 = 4.80 J
for x=0.08m from equilibrium, potential energy is
Ep = kx^2 / 2 = 667.29 * 0.08^2 / 2 = 2.13 J
therefore kinetic energy is
Ek = E - Ep = 4.80 - 2.13 = 2.67 J
Velocity in that point is
v = sqrt (2Ek/m) = sqrt (2*2.67/0.47) = 3.37 m/s

(c)
Determine the total energy of the system.
E = Ekmax = m vmax^2 / 2
E = 0.47 * 4.52^2 / 2 = 4.80 J

(d) Determine the total energy of the system. J (d) If the amplitude of oscillation were increased by a factor of 3.3, by what factor does the total energy change?
total energy is a function of the square of the amplitude. So, if A is increased by a factor of 3.3, then the energy factor is

Ef = 3.3^2 = 10.89


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