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/s^{2}

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

______s

(h) What is its acceleration at this time?

______m/s^{2}

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/s ^{2}**

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/sec ^{2}.**

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