Question 348: Oscillations

Question

A mass '$$\mathrm{m}_1$$' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released, mass performs S.H.M. of period '$$\mathrm{T}_1$$'. If the mass is increased by '$$\mathrm{m}_2$$', the time period becomes '$$\mathrm{T}_2$$'. The ratio $$\frac{\mathrm{m}_2}{\mathrm{~m}_1}$$ is

Accepted Answer

$$
\begin{aligned}
& 	ext { } \
& T_1=2 \pi \sqrt{\frac{m_1}{k}}, T_2=2 \pi \sqrt{\frac{m_1+m_2}{k}} \
& 	herefore \frac{T_2}{T_1}=\sqrt{\frac{m_1+m_2}{m_1}} \
& 	herefore \frac{T_2^2}{T_1^2}=\frac{m_1+m_2}{m_1} \
& 	herefore \frac{T_2^2-T_1^2}{T_1^2}=\frac{m_2}{m_1}
\end{aligned}
$$

Answer

$$\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2}$$

Answer

$$\frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{~T}_1}$$

Answer

$$\frac{\mathrm{T}_2^2-\mathrm{T}_1^2}{\mathrm{~T}_1^2}$$

Answer

$$\frac{T_1^2-T_2^2}{T_1^2}$$

Oscillations

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