Question

(II) At t =0, an 885-9 mass at rest on the end of a horizontal Spring (K: 184 N/m) is struck by a hammer which gives it aninitial speed of 226 m/s. Determine (a) the period and frequency ofthe motion, (b) the amplitude, (c) the maximumacceleration, (d) the total energy, and (e) the kinetic energy when x =O.4O A where A is the amplitude.

Answer 1

A. T = 0.4358s and f = 2.29hz

B. A = 15.67m

C. amax = 3258.71m/s

D. amax = 22601J

E. Ek = 3616.16J

Step-by-step explanation:

A. The period of the motion, T = 2pi*(sqrt(m/k))

Where m is the mass of the body in motion = 885g = 0.885kg

k = the spring constant = 184N/m2

T = 2pi*(sqrt(0.885/184))

= 0.4358s

Frequency of the motion, f = 1/T

T = 0.4358s

f = 2.2949hz

B. Maximum speed, Vmax = A*(sqrt(k/m))

Where A = amplitude of the motion

Making amplitude subject of formula,

A = Vmax(sqrt(m/k))

= 226*(sqrt(0.885/184))

= 15.6739m

C. Maximum acceleration, amax = A*(k/m)

= 15.6739*(184/0.885)

= 3258.71m/s

D. Total energy, Etotal = 1/2*(m * Vmax)2

= 1/2 * 0.885 * (226)2

= 22601J

E. Kinetic energy, Ek = Etotal - mechanical energy

Ek = 1/2*(k*A2) - 1/2*(k*x2)

Where x = 0.40A

Ek = 1/2*((k*A2) - (k*0.40A)2)

= 1/2*k*A2*(1 - 0.16)

= 1/2*k*A2*0.16

But 1/2*k*A2 = 22601J

Therefore, Ek = 22601*0.16

= 3616.16J