Mass on Ideal Spring A mass of M = .114 kg is hung from a spring with spring con

Question

Mass on Ideal Spring A mass of M = .114 kg is hung from a spring with spring constant k = 1.53 N/rn. The mass is displaced a distance xmax = -0.045 m and let go at time t = 0. Determine the spring's period of oscillation T and angular frequency omega. How long does it take take for the mass to move from xo = -xmax to the equilibrium? Is this time equal to T/4? Determine the mass's position, velocity and acceleration after 1 second. Determine the mass's position, velocity and acceleration after 10 seconds.

Explanation / Answer

Q1.

part a:

angular frequency=sqrt(k/m)=sqrt(1.53/0.114)=3.663 rad/s

then period of oscillation=2*pi/angular frequency=1.7153 seconds

b)the displacement of the mass at any time t can be modelled as

x=-0.045*cos(3.663*t)

then at equilirbrium point x=0

==>3.663*t=pi/2

==>t=0.4288 seconds

and yes, this time is equal to T/4

part c:

after 1 second, x(t)=-0.045*cos(3.663*1)=0.039 m

x(t)=-0.045*cos(3.663*t)

==>veloicty=v(t)=dx/dt=(-0.045)*(-3.663)*sin(3.663*t)

=0.1648*sin(3.663*t)

hence at t=1 second, veloicty=0.1648*sin(3.663*1)=-0.082 m/s

acceleration=dv/dt=0.1648*3.663*cos(3.663*t)

=0.60367*cos(3.663*t)

hence at t=1, acceleration=0.60367*cos(3.663*1)=-0.5234 m/s^2

part d:

at t=10:

position=x(t)=-0.045*cos(3.663*10)=-0.0216 m

veloicty=0.1648*sin(3.663*10)=-0.1445 m/s

acceleration=0.60367*cos(3.663*10)=0.2903 m/s^2

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