Question 313: Oscillations

Question

The upper end of the spring is fixed and a mass '$$m$$' is attached to its lower end. When mass is slightly pulled down and released, it oscillates with time period 3 second. If mass '$$\mathrm{m}$$' is increased by $$1 \mathrm{~kg}$$, the time period becomes 5 second. The value of '$$\mathrm{m}$$' is (mass of spring is negligible)

Accepted Answer

The formula for the time period of a spring mass system is $$T=2 \pi \sqrt{\frac{m}{k}}$$
For mass $$\mathrm{m}+1, \mathrm{~T}^{\prime}=2 \pi \sqrt{\frac{\mathrm{m}+1}{\mathrm{k}}}$$
Taking the ratio,
$$\begin{aligned}
\frac{\mathrm{T}}{\mathrm{T}^{\prime}} & =\frac{2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}}{2 \pi \sqrt{\frac{\mathrm{m}+1}{\mathrm{k}}}} \
\frac{\mathrm{T}}{\mathrm{T}^{\prime}} & =\sqrt{\frac{\mathrm{m}}{\mathrm{m}+1}} \
\frac{3}{5} & =\sqrt{\frac{\mathrm{m}}{\mathrm{m}+1}} \
\frac{9}{25} & =\frac{\mathrm{m}}{\mathrm{m}+1} \
	herefore \quad 9 \mathrm{~m} & +9=25 \mathrm{~m} \
	herefore \quad 16 \mathrm{~m} & =9 \
	herefore \quad \mathrm{m} & =\frac{9}{16} \mathrm{~kg}
\end{aligned}$$

Answer

$$\frac{3}{8} \mathrm{~kg}$$

Answer

$$\frac{5}{9} \mathrm{~kg}$$

Answer

$$\frac{8}{13} \mathrm{~kg}$$

Answer

$$\frac{9}{16} \mathrm{~kg}$$

Oscillations

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