Question

A 4.70-kg object on a frictionless horizontal surface is attached to one end of a horizontal spring that has a force constant k = 570 N/m. The spring is stretched 9.30 cm from equilibrium and released.

(a) What is the frequency of the motion?
_____Hz

(b) What is the period of the motion?
______s

(c) What is the amplitude of the motion?
______cm

(d) What is the maximum speed of the motion?
______m/s

(e) What is the maximum acceleration of the motion?
_____m/s²

(f) When does the object first reach its equilibrium position?
______s

(h) What is its acceleration at this time?
______m/s²

Answer 1

given

m = 4.70 - kg

force constant k = 570 N/m

The spring is stretched 9.30 cm = A

A = 0.093 m

a)

using the equation for the frequency of the motion is

f = (1/) ( k/m )^1/2

f = (1/) ( 570/4.7 )^1/2

f = 1.753 Hz

b )

T = 1/f

T = 1/1.753

T = 0.5704 sec

c )

amplitude is A = 9.3 cm

A = 0.093 m

d )

maximum speed is V_max = A

= A ( f )

= 0.093 X 2 X 3.14 X 1.753

= 1.023 m/s

e )

maximum acceleration

a_max = A ²

= A ( f )²

= 0.093 X ( 2 X 3.14 X 1.753)²  

= 11.27 m/s²

f )

t = T/4

t = 0.5704 / 4

t = 0.1426 sec

t = 142.6 msec

g )

at equilibrium position the acceleration is ZERO m/sec².