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A 25 kg object attached to an ideal spring with a force constant spring constant) of 15 N/m oscillates on a horizontal, frict
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Answer #1

spring mass system is

F= -kx

d. m E-ko dt2   (\frac{d^{2}x}{dt^{2}} = a)

\frac{d^{2}x}{dt^{2}} +\frac{k}{m}x=0

\frac{d^{2}x}{dt^{2}} +\frac{15}{2.5}x=0

\frac{d^{2}x}{dt^{2}} +6x=0

D^{2}+6= 0

auxiliary eqn is

m^{2}+6= 0

m^{2}=-6

m= \pm i\sqrt{6} or   m= 0\pm i\sqrt{6}

then solution is

x= e^{\alpha t }(C_{1}cos\beta t+C_{2}sin\beta t)

x= e^{0.t }(C_{1}cos\beta t+C_{2}sin\beta t)

x= (C_{1}cos\sqrt{6} t+C_{2}sin\sqrt{6}t)----------------------------------(1)

now initial condition

x(t=0) =8cm   

v (t=0)= 0 m/s ( b/c x at t=0 is  constant so v=0 at t=0 )

Differentiate  the eqn(1)

v=\frac{dx}{dt}= \sqrt{6}(-C_{1}Sin\sqrt{6} t+C_{2}Cos\sqrt{6}t)

using intial condtion

v(t=0)=0 m/s

0= \sqrt{6}C_{2}

C_{2}=0--------------------------(2)

and x(t=0)=8cm

from eqn (1)

8= C_{1}-------------------------(3)

put the eqn(2) and eqn (3) in eqn (1)

x=0.08Cos\sqrt{6}t

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