8/14/2019 University Of Sheffield - Structural Engineering Masters - Vibration Engineering Coursework 1 Solutions

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Vibration Engineering Coursework 1

Question 1

System 1

fn,1 = 4.5 Hz m = 6500 kg1 = fn,1 * 2 * = 28.27 rad / sec

1 = ( k1 / m1 )0.5

k1 = 5196285 N / m

Question 2

System 1

With the initial conditions : x(0) = 0.05 m and v(0) = 1m/s

The first 5 seconds of the displacement, velocity and the acceleration graphs are :

The absolute values of the velocities are times the absolute value ofthe displacements and the

absolute value of the accelerations are 2

times the absolute value of the displacements. This is

because the function of the motion is given by a sinusoidal function, and because the velocity is the

derivate of the displacements, and accelerations are the second derivate, the natural frequency

value comes out of the brackets of sinusoidal function, and because we have a natural frequency

greater than 1 rad / seconds, the results are in this way. Also the maximum displacement is

somewehere about 0.06 meters this is because of the 0.05 meters initial displacement and plus theinitial velocity that is applied to the system.

System 2

fn,1 = 0.2 Hz m = 6500 kg1 = fn,1* 2 * = 1.256 rad / sec

1 = ( k1 / m1 )0.5

k1 = 10254 N / m

8/14/2019 University Of Sheffield - Structural Engineering Masters - Vibration Engineering Coursework 1 Solutions

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The spring force, inertial force and the sum for the 5 five seconds are :

Since there is no external force acting, the only energy in the system is given by the initial

displacement and the velocity. Also because there is no damping force, i.e. energy loss, the total

energy in the system never changes.

8/14/2019 University Of Sheffield - Structural Engineering Masters - Vibration Engineering Coursework 1 Solutions

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Question 3

With the initial conditions : x(0) = 0.05 m and v(0) = -1m/s

Care must be taken to the x and y axiss when observing the two graphs of system 1 (on the left) and

the system 2 (on the right). The high frequency system completes one cycle in around 0.22 seconds,

whereas the low frequency system completes one cycle in five seconds.

There is a great difference between ( about x13 times ) the maximum displacements the two

different systems make. This is caused by the initial velocities applied to the systems. If only initial

displacements were to be applied, the maximum displacements the systems make would be both

same and equal to the initial displacement that is applied.

However, having in mind that the systems have the same mass, the same velocities the masses have

means that the same amount of Kinetic Energy the masses have. However, system 2 has a much

lower stiffness compared to system 1( which also causes the low frequency with the same mass

values), it is not hard to imagine, that it will make much more displacement.

When the maximum displacement happens, it is where the Kinetic Energy gets the lowest value, and

the Potential Energy gets the highest. Lower stiffness needs a larger displacement to be equal to the

initial Kinetic Energy.

Also, if only ( and equal ) initial displacements were to be applied, much more energy would be

needed for applying the initial condition for the high frequency system. This also verifies why they

would have the same amount of displacements if only initial displacements were to be applied, from

the energy point of view.