Answer:
Maximum force will be 29040 N
Explanation:
We have given mass of the piston m = 1.5 kg
Amplitude A = 10 cm = 0.1 m
Angular speed [tex]\omega =4200rpm=\frac{2\times 3.14\times 4200}{60}=440rad/sec[/tex]
We know that angular speed is given by [tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]440=\sqrt{\frac{k}{1.5}}[/tex]
Squaring both side
[tex]193600=\frac{k}{1.5}[/tex]
k = 290400 N/m
Now force is given by [tex]F=kA=290400\times 0.1=29040N[/tex]