Correct Answer - Option 2 : 1.6 kg
Concept:
The relation between time-period (T), angular velocity (ω) and frequency (f) is given by:
\(T=\frac{2π}{ω}=\frac{1}{f}\)
We know that for undamped vibration:
\(\omega=\sqrt{\frac{k}{m}}\)
Therefore
\(T=\frac{2π}{ω}=2\pi \sqrt{\frac{m}{k}}\)
Calculation:
Given:
T1 = 2 sec, m1 = m kg
T2 = 3 sec, m2 = (m + 2) kg.
\(T=2\pi \sqrt{\frac{m}{k}}\)
\(\frac{T_2}{T_1}= \sqrt{\frac{m_2}{k}}\times\sqrt{\frac{k}{m_1}}\)
\(\frac{3}{2}= \sqrt{\frac{m\;+\;2}{m}}\)
\(\frac{9}{4}= \frac{m\;+\;2}{m}\)
9m = 4m + 8
5m = 8
m = 1.6 kg