12. Which of the following statements about
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Refer to the diagram shown.
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N
By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m
When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.
If damping is ignored, the equation of motion is
F = m * acceleration
or
Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)
x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0
The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32
The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.
Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)
The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.
Explanation:
When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.
As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.
Since, the inlet volume flow is measured as the product of velocity with the area.
Inlet volume flow=Inlet velocity*Area*time
And the mass flow rate is
Mass flow rate in the inlet=density*area*inlet velocity*time
Mass flow rate in the outlet=density*area*outlet velocity*time
Since, the time and area is constant, the inlet and outlet will be same as
(Mass inlet)/(density*inlet velocity)=Area*Time
(Mass outlet)/(density*outlet velocity)=Area*Time
As the ratio of mass to density is termed as specific volume, then
(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)
Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity
As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity
is
So, the inlet velocity is 1.4035 m/s.
Then the inlet volume will be
As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s
So, the inlet volume is 0.019 m³/s.
Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.