Question

A 0.520 kg mass suspended from a spring oscillates with a period of 1.50 s. how much mass must be added to the object to change the period to 2.05 s?

Answer 1

$Given:\\m_1=0.520kg\\T_1=1.50s\\T_2=2.05s\\\\Find:\\\Delta m=?\\\\Solution:\\\\\Delta m=m_2-m_1\\\\T=2 \pi \sqrt{ \frac{m}{k} } \Rightarrow 2 \pi = \frac{T}{\sqrt{ \frac{m}{k} }} \\\\\frac{T_1}{\sqrt{ \frac{m_1}{k} }} =\frac{T_2}{\sqrt{ \frac{m_2}{k} }} \\\\T_1\sqrt{ \frac{m_2}{k} }} =T_2\sqrt{ \frac{m_1}{k} }} \\\\\frac{T_1^2m_2}{k} =\frac{T_2^2m_1}{k} \\\\T_1^2m_2=T_2^2m_1\Rightarrow m_2= \frac{T_2^2m_1}{T_1^2} \\\\\Delta m=\frac{T_2^2m_1}{T_1^2} -m_1=m_1(\frac{T_2^2}{T_1^2} -1)$

$\\\\\Delta m=0.520kg( \frac{(2.05s)^2}{(1.50s)^2} -1)\approx 0.4512kg$