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SaRaH...
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We have been given this as a sample exam question.
For a car store a small model car (75kg) is hung from the ceiling. The model is hanging on an elastic spring with force constant 1500 N/m. The model is oscillating 10 times per second with an amplitude of 0.02 m. (Assume a zero-phase shift).
a) What is the initial extension of the spring?
b)Determine the velocity when the model passes through the equilibrium point.
c) What is the velocity when the model is -0.01 m from equilibrium and what direction has the velocity?mass = 75kg
force/weight = mg = 735.75N
k = 1500N
Amplitude = 0.02m
frequency = 10Hz
For part a) I have calculated, using the expression F = kx, that the extension of the spring (x) is 0.4905m.
For part b) I know that the maximum velocity will occur at equilibrium as this is where potential energy is 0 and there is maximum kinetic energy. Using the equation v=ωAcos(ωt) and the fact that the max. value of cos(ωt) = I have found that maximum velocity is v=ωA.
Should this be calculated using ω=2πf so that the solution is:
Vmax = ωA = 2πfA = 2π(10)(0.02) = 1.26 m/s
or should ω be calculated as ω=sqrt(k/m) so the solution is:
Vmax = ωA = sqrt(k/m)A = sqrt(1500/75)(0.02) = 0.089 m/s
Should these not give equivalent answers?
Also for part c) I have calculated the following:
x(t) = -0.01m v(t) = ?
-0.01 = Asin(ωt)
solving this for t gives t = 1/ωcos(sin^-1(-0.01/A))
Substituting this into the equation for velocity v=ωAcos(ωt) gives v = ωAcos(sin^-1(-0.01/A))
However to finish this I have the same problem as above in b).There is probably just something simple that I am overlooking and any help with this would be very much appreciated.
Thank You,
Sarah. :)
For a car store a small model car (75kg) is hung from the ceiling. The model is hanging on an elastic spring with force constant 1500 N/m. The model is oscillating 10 times per second with an amplitude of 0.02 m. (Assume a zero-phase shift).
a) What is the initial extension of the spring?
b)Determine the velocity when the model passes through the equilibrium point.
c) What is the velocity when the model is -0.01 m from equilibrium and what direction has the velocity?mass = 75kg
force/weight = mg = 735.75N
k = 1500N
Amplitude = 0.02m
frequency = 10Hz
For part a) I have calculated, using the expression F = kx, that the extension of the spring (x) is 0.4905m.
For part b) I know that the maximum velocity will occur at equilibrium as this is where potential energy is 0 and there is maximum kinetic energy. Using the equation v=ωAcos(ωt) and the fact that the max. value of cos(ωt) = I have found that maximum velocity is v=ωA.
Should this be calculated using ω=2πf so that the solution is:
Vmax = ωA = 2πfA = 2π(10)(0.02) = 1.26 m/s
or should ω be calculated as ω=sqrt(k/m) so the solution is:
Vmax = ωA = sqrt(k/m)A = sqrt(1500/75)(0.02) = 0.089 m/s
Should these not give equivalent answers?
Also for part c) I have calculated the following:
x(t) = -0.01m v(t) = ?
-0.01 = Asin(ωt)
solving this for t gives t = 1/ωcos(sin^-1(-0.01/A))
Substituting this into the equation for velocity v=ωAcos(ωt) gives v = ωAcos(sin^-1(-0.01/A))
However to finish this I have the same problem as above in b).There is probably just something simple that I am overlooking and any help with this would be very much appreciated.
Thank You,
Sarah. :)