Câu hỏi:

Ta có tần số $f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$.

• Khi gắn $m_1 + m_2$: $f_1 = 2 = \frac{1}{2\pi}\sqrt{\frac{k}{m_1+m_2}}$ (1)
• Khi gắn $m_1 - m_2$: $f_2 = 4 = \frac{1}{2\pi}\sqrt{\frac{k}{m_1-m_2}}$ (2)

Chia (1) cho (2) ta được: $\frac{1}{2} = \sqrt{\frac{m_1-m_2}{m_1+m_2}} \Rightarrow \frac{1}{4} = \frac{m_1-m_2}{m_1+m_2} \Rightarrow m_1+m_2 = 4m_1 - 4m_2 \Rightarrow 5m_2 = 3m_1 \Rightarrow m_1 = \frac{5}{3}m_2$.
Thay $m_1 = \frac{5}{3}m_2$ vào (1) ta có: $2 = \frac{1}{2\pi}\sqrt{\frac{k}{\frac{5}{3}m_2 + m_2}} = \frac{1}{2\pi}\sqrt{\frac{k}{\frac{8}{3}m_2}} \Rightarrow 4 = \frac{1}{4\pi^2} \frac{3k}{8m_2} \Rightarrow \frac{k}{m_2} = \frac{128\pi^2}{3}$.
Khi gắn $m_1$: $f_{m_1} = \frac{1}{2\pi}\sqrt{\frac{k}{m_1}} = \frac{1}{2\pi}\sqrt{\frac{k}{\frac{5}{3}m_2}} = \frac{1}{2\pi}\sqrt{\frac{3k}{5m_2}} = \frac{1}{2\pi}\sqrt{\frac{3}{5} \cdot \frac{128\pi^2}{3}} = \frac{1}{2\pi}\sqrt{\frac{128\pi^2}{5}} = \sqrt{\frac{32}{5}} \approx 2.5298$ Hz. $T_{m_1} = \frac{1}{f_{m_1}} \approx 0.3953$ s.
Khi gắn $m_2$: $f_{m_2} = \frac{1}{2\pi}\sqrt{\frac{k}{m_2}} = \frac{1}{2\pi}\sqrt{\frac{128\pi^2}{3}} = \sqrt{\frac{32}{3}} \approx 3.266$ Hz. $T_{m_2} = \frac{1}{f_{m_2}} \approx 0.3062$ s.