Question #216076

A machine tool has a total mass, m of 150 kg and it is supported on springs with an equivalent stiffness, k = 3375 N/m. Determine:

1.1. The static deflection due to the tool. (1)

1.2. The angular velocity and period of the simple harmonic motion of the tool. (2)

1.3. The amplitude of the simple harmonic motion of the tool if initial conditions are x(0) = 0.08m and v(0) = 0 m/s (4)

1.4. The maximum velocity and maximum acceleration of the motion. (4)


Expert's answer

1)


x=150(9.8)3375=0.4356 mx=\frac{150(9.8)}{3375}=0.4356\ m

2)


ω=km=3375150=4.743radsT=2π1503375=1.325 s\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{3375}{150}}=4.743\frac{rad}{s}\\T=2\pi\sqrt{\frac{150}{3375}}=1.325\ s

3)


A=0.08 mA=0.08\ m

4)


V=0.08(7.743)=0.6194msa=0.08(7.743)2=4.796ms2V=0.08(7.743)=0.6194\frac{m}{s}\\a=0.08(7.743)^2=4.796\frac{m}{s^2}


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