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When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, its frequency is 12Hz. When another object of mass m2 is hung on the spring along with m1, the frequency of the motion is 4Hz. Find the ratio m2/m1 of the masses.

Respuesta :

Answer:

Explanation:

Given

When [tex]m_1[/tex] mass is attached to the spring its frequency is [tex]f_1=12\ Hz[/tex]

when another mass [tex]m_2[/tex] is attached to [tex]m_1[/tex] , frequency changes to [tex]f_2=4\ Hz[/tex]

frequency of spring mass system is given by

[tex]f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}[/tex]

for [tex]m_1[/tex]

[tex]f_1=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}[/tex]

[tex]12=\frac{1}{2\pi }\sqrt{\frac{k}{m_1}}-----1[/tex]

for [tex]m_1[/tex] and [tex]m_2[/tex]

[tex]f_2=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}[/tex]

[tex]4=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}----2[/tex]

divide 1 and 2

[tex]3=\sqrt{\frac{m_1+m_2}{m_1}}[/tex]

squaring

[tex]9=1+\frac{m_2}{m_1}[/tex]

[tex]\frac{m_2}{m_1}=8[/tex]

   

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