Ablock of mass 0.75 kg is suspended from a spring having a stiffness of 150 n/m. the block is displaced downwards from its equilibrium position through a distance of 3 cm with an upward velocity of 2 cm/sec. determine: a)- the natural frequency b)-the period of oscillation c)-the maximum velocity d)-the phase angle e)-the maximum acceleration

a)f=2.25 Hz

b)Time period T=.144 s

c)tex]V_{max}[/tex]=0.42 m/s

d)Phase angle Ф=87.3°

e)

Explanation:

a)

Natural frequency

=14.14 rad/s

w=2πf

⇒f=2.25 Hz

b) Time period

T=

 Time period T=.144 s

c)Displacement equation

Boundary condition

t=o,x=0.03 m

t=0,v=.02m/s   , V=

Now by using these above conditions

A=0.03,B=0.0014

x=0.03 cos14.14 t+0.0014 sin14.14 t

⇒x=0.03003sin(14.14t+87.3)

=0.42 m/s

d)

Phase angle Ф=87.3°

e)

Maximum acceleration

=6.0041

answer:

d. attempt to find the owner, then leave your information if you cannot find them.

explanation:

d. represents both (b) and (c).

more logical as well.

b and c are to complicated.

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