At t=0, an 830-g mass at rest on the end of a horizontal spring (k = 169 N/m) is struck by a hammer which gives it an initial speed of 2.32 m/s.

Determine (a) the period and frequency of the motion, (b) the amplitude, (c) the maximum acceleration, (d) the total energy, and (e) the kinetic energy when

x = 0.40 A, where A is the amplitude.

Answer:a) T=0.44seconds and f=2.29Hz

b) A=0.1626m

c) amax=33.11m/s^2

d) E=2.2337J

e) KE=1.8732J

Explanation:

Period of the motion:

T=2π√m/k

T=period, m=mass of the body, k=spring constant

Converting mass from gram to kilogram:

M=830g/1000=0.83kg

T=2π√0.83/169=0.44038seconds

T=0.44secs (approximately)

Frequency of the motion:

F=1/T

Where T=0.44038seconds

F=1/0.44038s=2.27Hz

b) maximum speed:

Vmax=A√k/m

Where A=amplitude, m=mass of the body, k=the spring constant

Making A the subject of the formula:

A=Vmax√m/k

Where, speed (V) = 2.32m/s as Vmax and 0.83kg for m and 169N/m for k.

A = (2.32m/s) √ (0.83kg) / (169N/m) = 0.1626m

c) Maximum acceleration:

amax = A (k/m)

Therefore, A=

(0.1626m) (169N/m) / 0.83kg)

A=33.11m/s^2

d) Total Energy:

E=1/2MVmax

Where, M=0.83kg and Vmax=2.32m/s

E=1/2 * (0.83kg) * (2.32m/s) ^2

E=2.2337J

e) From the law of conservation of mass:

1/2kA^2=1/2Kx^2+KE

Where x=0.04A

KE=1/2kA^2-1/2kx^2

KE=1/2kA^2 (1-0.4^2)

KE=Etotal*0.84

Where, Etotal=2.23J

KE=2.23*0.84=1.8732J