#$&*
course Phy 201
12/3 5 pmBelow I am going to post of the completed questions from the final. They are all completed minus the last one and that one I have worked, but I don't know where to go from the point i'm at.
Problem Number 12
A simple harmonic oscillator with mass 2.33 kg and restoring force constant 320 N/m is released from rest at a displacement of .49 meters from its equilibrium position.
• What is its equation of motion?
• According to the corresponding acceleration function what will be the acceleration of the oscillator at clock time t = .1714 sec?
• What will be its position at this clock time?
• What is the force on the oscillator at this position?
• Does this force result in the acceleration you just calculated?
F=-kx=-320Nm(.49m)=-156.8 N
Omega=sqrt(k/m)=sqrt(320 Nm/2.33 kg)=1.77 rad/s
The equation of motion is x = A cos(omega t) = .49 meters cos(1.77 rad/s * t).
At t = 0 this gives you x = .49 meters cos(0) = .49 meters, so if you use the cosine function you can assume that the object was released at t = 0.
Theta=omega*t=1.77rad/s*.1714 s=.30rad
Y position .49 m sin .30 rad=.0026 m
Force at this position =-320Nm(.026m)=8.32 N
a=`dv/`dt=1.77 rad/s/.1714s=10.3 rad/s^2
1.77 rad/s is an angular velocity, not a change in velocity.
`dv / `dt would in any case give you only an average rate of change. Since acceleration isn't constant in this situation, that won't help.
You could however divide the force by the mass. This would give you an acceleration.
F=ma=2.33kg *10.3rad/s= 24N No it does not result in the acceleration I just calculated.
10.3 rad/s, or 10.3 rad/s^2, is not an acceleration. The latter is an angular acceleration, but the quantity wasn't calculated in a way that is consistent with this situation.
If you use the cosine function instead of the sine, and divide the force by the mass, you should get an acceleration that agrees with the given information.
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Problem Number 9
A simple harmonic oscillator is subjected to a net restoring force F = - 500 N/m * x at displacement x from equilibrium. It is observed to undergo 31 complete cycles of motion in 32 seconds. What is its mass?
31 cycles/32 seconds=.97 cycle/s
2pi rad/.97 s=6.5 rad/s
6.5 rad/s=sqrt(-500Nm/mass) Square both sides
6.5 rad^2/s^2=-500Nm/mass
.6.5 rad^2/s^2*mass=-500Nm Divide by angular frequency
Mass=76.9 kg
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Problem Number 10
What is the centripetal acceleration of a satellite orbiting at a radius of 18600 km from the center a certain planet if it is moving at 16000 m/s in that orbit? What is its orbital period (i.e., how long does it take to complete an orbit)?
Centripetal Acceleration=v^2/r (16000m/s)^2/18600000 m=13.76 m/s^2
18600 km/16000m/s=1162.5 s or 19.375 minutes
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Problem Number 7
A simple pendululm of length 3 meters and mass .51 kg is pulled back a distance of .239 meters in the horizontal direction from its equilibrium position, which also raises it slightly. What is its gravitational potential energy increase?
F=mg/L*`dx .51kg(9.8m/s^2)/3m*.239 m=.4 N
Is the force pulling the pendulum back to equilibrium the same as the increase of gravitational force? Or would be PE=mg*h=.51 kg*9.8 m/s^2*.239 m=1.1945 J which is the change in PE from equilibrium?
That would be right if .239 m was the change in vertical position. However .239 m is the pullback, which is not the change in its vertical position. The .239 m pullback is mostly in the horizontal direction.
To get the change in vertical coordinate you could first find the vertical leg of a triangle with hypotenuse 3 meters and horizontal leg .239 meters. Subtract this from the 3 meter length and you would have the change in vertical position, which you could then multiply by the weight of the pendulum to get the change in PE.
Or you could use PE = 1/2 k x^2.
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Problem Number 2
A simple harmonic oscillator is subjected to a net restoring force F = - 80 N/m * x at displacement x from equilibrium. It is observed to undergo simple harmonic motion with a frequency of 1.9 cycles / second. What is its mass?
2 pi rad*1.9 cycle/s=11.9 rad/s
11.9 rad/s=sqrt(-80Nm/mass) To solve, first square both sides
141 rad^2/s^2=-80Nm/mass Multiply by mass
141 rad^2/s^2* mass=-80Nm Then divide by 141 rad^2/s^2
Mass=.57 kg
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Problem Number 2
A simple pendulum has a length of 2.45 meters and a mass of .34 kg. It is given a KE of .187 Joules at a point .1641 meters from equilibrium. What will be its maximum displacement from equilibrium?
Looking for x +.1641 m
KE=1/2 mv^2
.187 J=1/2(.34 kg)(v^2)
.187 J/.17 kg=v^2
1.1 m^2/s^2=v^2
v=1.049 m/s
K=(1.049m/s)^2*.34=.374Nm
F=-kx---I am not sure where to go with this problem.
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You know the KE at the .164 m point, and you can find the PE at that point, which is 1/2 k x^2 with x = .164 m and k = m g / L.
The ideal pendulum doesn't lose mechanical energy, so the total energy at the .164 m point is the total energy at every point. So you can just add the KE and the PE at the .164 m point, giving you the total energy.
At maximum displacement x = A the KE is zero, so the PE is equal to the total energy. Since the PE at this point is 1/2 k A^2, you can set 1/2 k A^2 equal to the total energy and solve for A.
Good questions. You might well have questions about my responses, or about other problems. Your questions are encouraged.