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Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1, the spring constant is 2k, the mass is m, and the spring oscillates with amplitude d. In CASE 2, the spring constant is 2k, the mass is 2m, and the spring oscillates with amplitude 2d.

Respuesta :

Manetho

Answer:

[tex]\frac{F_1}{F_2} =\frac{1}{2}[/tex]

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

[tex]F_1=2k\times d[/tex]

case 2:

[tex]F_2=2k\times 2d[/tex]

therefore, [tex]\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}[/tex]

[tex]\frac{F_1}{F_2} =\frac{1}{2}[/tex]

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