Differential Equations problem

Question

As the engineer in charge of the calibration software, you are to create the correct algorithm to analyze

the desk on which the printer is. Let us focus on some part of this. Assume the machine is the object

producing an external vibration on the desk, a pure sine function, ranging from 1 to 10 Hz, with an

amplitude kept constant at 1 mm (for sake of simplicity). Consider the following information for the

desk, no damping considered, and assume the initial condition are x(0) = 0 and x’(0) = 0.

M = 70 kg ; k = 7 kN /m

a) Write a differential equation that would work for all frequencies in the range from 1 to 10 Hz.

b) What is the natural frequency of the desk?

c) Solve the differential equation found in a) for the case when the applied frequency is different

than the one found in b).

d) Do a graph of the amplitude of the response as a function of the applied frequency of 1.55, 1.58,

1.585, 1.6, 1.61, and 1.65 Hz. What do you understand from this graph? (You don’t need to be

exact here, we are looking for the general features/trend of the graph; you can neglect some

components of the exact amplitude)

Clearly state all your hypotheses!

Answer 1

Hi Soyeb,

Please see detailed response below.

a) The differential equation describing the motion of the desk subjected to an external vibration is derived as follows. Considering the desk as a mass-spring system with base excitation, the equation of motion is:

x′′+100x=0.1sin⁡(ωt)x′′+100x=0.1sin(ωt)

where ωω is the angular frequency of the applied vibration.

b) The natural frequency of the desk is calculated using the formula ωn=kMωn​=Mk​​. Converting this to Hz:

fn=ωn2π=102π≈1.59 Hzfn​=2πωn​​=2π10​≈1.59 Hz

Thus, the natural frequency is 1.59 Hz1.59 Hz​.

c) Solving the differential equation for the case when the applied frequency ωω is different from the natural frequency ωnωn​, we use the method of undetermined coefficients. The solution is:

x(t)=0.1100−ω2(sin⁡(ωt)−ω10sin⁡(10t))x(t)=100−ω20.1​(sin(ωt)−10ω​sin(10t))

d) The amplitude of the response as a function of the applied frequency shows a resonance peak near the natural frequency. The amplitudes calculated for the given frequencies are approximately:

1. 1.55 Hz: 19.3 mm
2. 1.58 Hz: 69.4 mm
3. 1.585 Hz: 125 mm
4. 1.6 Hz: 94.3 mm
5. 1.61 Hz: 45.2 mm
6. 1.65 Hz: 13.7 mm

The graph illustrates a sharp increase in amplitude as the applied frequency approaches the natural frequency (1.59 Hz), peaking near 1.585 Hz, and then decreasing as the frequency moves away. This demonstrates the resonance phenomenon in an undamped system, where amplitudes become very large near the natural frequency.

Final Answers

a) \boxed{x'' + 100x = 0.1 \sin(\omega t)}

b) \boxed{1.59 \text{ Hz}}

c) The solution is:

x(t)=0.1100−ω2(sin⁡(ωt)−ω10sin⁡(10t))x(t)=100−ω20.1​(sin(ωt)−10ω​sin(10t))

d) The graph shows a resonance peak near the natural frequency, highlighting the dramatic increase in amplitude as the applied frequency approaches the natural frequency.