Question 293: Oscillations

Question

A spring has a certain mass suspended from it and its period of vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $\mathrm{T}_2$. The ratio of $T_2 / T_1$ is

Accepted Answer

$$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} \Rightarrow \mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{~K}}} \
	herefore \quad & \frac{\mathrm{~T}_2}{\mathrm{~T}_1}=\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}\quad	ext{.... (i)}
\end{aligned}$$
Let K be the spring constant of the spring cut in half.
For series combination,
$$K_1=\frac{K 	imes K}{K+K}=\frac{K}{2}$$
For second case, the mass is hung to only one half of the spring,
$$\mathrm{K}_2=\mathrm{K}$$
Substituting in $\mathrm{K}_1$ and $\mathrm{K}_2$ in (i),
$$\frac{T_2}{T_1}=\sqrt{\frac{K / 2}{K}}=\frac{1}{\sqrt{2}}$$

Answer

$1: 2$

Answer

$1: \sqrt{2}$

Answer

$\sqrt{2}: 1$

Answer

$2: 1$

Oscillations

A spring has a certain mass suspended from it and its period of vertical oscillations is T_1. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now \mathrm{T}_2. The ratio of T_2 / T_1 is

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