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{ (m)/(k) } \Rightarrow 2 \pi = \frac{T}{\sqrt{ (m)/(k) }} \\\\\frac{T_1}{\sqrt{ (m_1)/(k) }} =\frac{T_2}{\sqrt{ (m_2)/(k) }} \\\\T_1\sqrt{ (m_2)/(k) }} =T_2\sqrt{ (m_1)/(k) }} \\\\(T_1^2m_2)/(k) =(T_2^2m_1)/(k) \\\\T_1^2m_2=T_2^2m_1\Rightarrow m_2= (T_2^2m_1)/(T_1^2) \\\\\Delta m=(T_2^2m_1)/(T_1^2) -m_1=m_1((T_2^2)/(T_1^2) -1)`


`\\\\\Delta m=0.520kg( ((2.05s)^2)/((1.50s)^2) -1)\approx 0.4512kg`