Answer:
[tex]m_2/m_1=9.745[/tex]
The ratio m_2/m_1 of the masses is 9.745
Explanation:
Formula for frequency when mass m_1 is hung on the spring:
[tex]f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}[/tex]
where:
k is the spring constant
Formula for frequency when mass m_2 is hung on the spring along with m_1:
[tex]f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}[/tex]
where:
k is the spring constant.
In order to find ratio m_2/m_1, Divide the above equations:
[tex]\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}[/tex]
On Solving the above equation:
[tex]\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745[/tex]
The ratio m_2/m_1 of the masses is 9.745
