Question

A 0.49 kg mass at the end of a spring vibrates 8.0 times per second with an amplitude of 0.10 m.

(a) Determine the velocity when it passes the equilibrium point.

m/s

(b) Determine the velocity when it is 0.10 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 3.8, by what factor does the total energy change?

Answer 1

given values

f = 8.0 Hz

A = 0.1 m

m = 0.49 kg

a)

w = 2*pi*f = 2*(22/7)*8 = 50.286 rad/s²

v = w*A = 50.286*0.1 = 5.0286 m/s²

b)

the velocity when it is 0.10 m from equilibrium = w*sqrt(A² - x²)

                                                                   = 50.286*sqrt(0.1² - 0.1²) = 0

c)

Total energy = 1/2*mass*w²*A²

E = 0.5*0.49*(50.286)^2*(0.1)^2 = 6.195 J

d)

Anew = 3.8*A

so Total energy(new) = (3.8)²*E

                               = 14.44*E