Question

When an object of mass m1 is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of 11.8 Hz. When another object of mass m2 is hung on the spring along with the first object, the frequency of the motion is 3.60 Hz. Find the ratio m2/m1 of the masses.

Answer 1

Answer:

$m_2/m_1=9.745$

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:

$f_1=\frac{1}{2\pi}\sqrt{\frac{k}{m_1}}$

where:

k is the spring constant

Formula for frequency when mass m_2 is hung on the spring along with m_1:

$f_2=\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}$

where:

k is the spring constant.

In order to find ratio m_2/m_1, Divide the above equations:

$\frac{f_1}{f_2} =\frac{ \frac{1}{2\pi}\sqrt{\frac{k}{m_1}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}}$

On Solving the above equation:

$\frac{f_1}{f_2} =\frac{\sqrt{\frac{k}{m_1}}}{\sqrt{\frac{k}{m_1+m_2}}}\\(\frac{f_1}{f_2})^{2} =\frac{m_1+m_2}{m_1} \\(\frac{11.8}{3.60})^2= \frac{m_1+m_2}{m_1} \\10.745=\frac{m_1+m_2}{m_1}\\10.745m_1=m_1+m_2\\m_2=10.745m_1-1m_1\\m_2/m_1=9.745$

The ratio m_2/m_1 of the masses is 9.745