Let x be the displacement of the mass from its equilibrium position, and let t r

Question

Let x be the displacement of the mass from its equilibrium position, and let t represent time. Assume that t = -0s , the displacement is x = +2.50 cm and th x-component of velocity is Vx = -0.6928 m/s
(a) Calculate ?, the angular frequency of the simple harmonic motion. (b) Determine k, the spring constant of the spring (c) Calculate the total mechanical energy E of this mass-spring system (d) Calculate Xm, the displacement amplitude of the oscillatory motion (e) Calculate Vmax x , the velocity amplitude ( i.e., maximum speed ) of the mass as it oscillates The displacement at time t is given by: x(t) = Acos( ?t+ ?0 ), where A is the displacement amplitude and ?0 is the phase angle (f) Given that X = +2.50cm and Vx= -0.6928 m/s when t= 0s , calculate ?, the phase angle in radians T 0.3927s nm 7m -A x=0 +A

Explanation / Answer

a)angular frequency=2*pi/time period=2*pi/0.3927=16 rad/sec

b)as we know, angular frequency=sqrt(k/m)

==>16=sqrt(k/0.25)

==>k=16^2*0.25=64 N/m

c)as any point, total energy is conserved

hence total energy=spring potential energy+kinetic energy of the block

=0.5*k*x^2+0.5*m*v^2

at t=0, x=2.5 cm=0.025 m

v=-0.6928 m/s

then total energy=0.5*64*0.025^2+0.5*0.25*0.6928^2

=0.08 J

part d:

at maximum displacement, speed=0

hence total energy=potential energy of the spring

hence 0.08=0.5*64*xm^2

==>xm=sqrt(0.08/(0.5*64))=0.05 m=5 cm

part e:

x(t)=0.05*cos(16*t+phi)

at t=0, x(t)=2.5 cm=.025 m

==>0.025=0.05*cos(16*0+phi)

==>cos(phi)=0.5

==>phi=pi/3

hence x(t)=0.05*cos(16t+pi/3)

then v(t)=-0.05*16*sin(16*t+pi/3)

then vmax=16*0.05=0.8 m/s

part f:

phi=pi/3 radians
as calculated in part e

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