Question #225768

Q3. A mass of 5kg is hung from a tension spring causing an increase in the length of the spring by 18mm. Calculate the following:


a) The period of the vibration.


b) The frequency of the vibration.


c) Determine the maximum velocity and maximum acceleration of the mass when it is pulled down a further 15 mm.


d) If a 15 kg machine is suspended from this spring and the machine exerts an out-of-balance force of 10 N at 5 Hz. Determine the amplitude of the steady state vibration.


Expert's answer

a)


mg=kx→k=mg/x=5⋅9.81/0.018=2725 (N/m)mg=kx\to k=mg/x=5\cdot9.81/0.018=2725\ (N/m)


T=2πm/k=2⋅3.14⋅5/2725=0.269 (s)T=2\pi\sqrt{m/k}=2\cdot3.14\cdot\sqrt{5/2725}=0.269\ (s)


b)


f=1/T=1/0.269=3.72 (Hz)f=1/T=1/0.269=3.72\ (Hz)


c)


vmax=Aω=A⋅2πf=0.015⋅2⋅3.14⋅3.72=0.35 (m/s)v_{max}=A\omega=A\cdot2\pi f=0.015\cdot2\cdot3.14\cdot3.72=0.35\ (m/s)


amax=Aω2=A⋅(2πf)2=0.015⋅(2⋅3.14⋅3.72)2=8.2 (m/s2)a_{max}=A\omega^2=A\cdot(2\pi f)^2=0.015\cdot(2\cdot3.14\cdot3.72)^2=8.2\ (m/s^2)


d)


A=Fm⋅(ω02−ω2)2=Fm⋅((2πf0)2−(2πf)2)2=1015⋅((2⋅3.14⋅3.72)2−(2⋅3.14⋅5)2)2=0.0015 (m)A=\frac{F}{m\cdot\sqrt{(\omega^2_0-\omega^2)^2}}=\frac{F}{m\cdot\sqrt{((2\pi f_0)^2-(2\pi f)^2)^2}}=\frac{10}{15\cdot\sqrt{({(2\cdot3.14\cdot 3.72)^2-(2\cdot3.14\cdot5)^2})^2}}=0.0015\ (m)













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