Question 337: Oscillations

Question

A particle connected to the end of a spring executes S.H.M. with period '$$T_1$$'. While the corresponding period for another spring is '$$\mathrm{T}_2$$'. If the period of oscillation with two springs in series is 'T', then

Accepted Answer

$$T_1=2 \pi \sqrt{\frac{m}{k_1}} \quad T_2=2 \pi \sqrt{\frac{m}{k_2}}$$
When the two springs are connected in series, the effective spring constant is given by
$$\frac{1}{\mathrm{k}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}$$
When connected in series
$$\begin{aligned}
\mathrm{T} & =2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \
	herefore \mathrm{T}^2 & =4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}}=4 \pi^2 \mathrm{~m}\left[\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}ight]=4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_1}+4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_2} \
& =\mathrm{T}_1^2+\mathrm{T}_2^2 \
	herefore \mathrm{T} & =\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}
\end{aligned}$$

Answer

$$\mathrm{T}=\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}$$

Answer

$$\mathrm{T}=\sqrt{\mathrm{T}_2^2-\mathrm{T}_1^2}$$

Answer

$$\mathrm{T=T_1+T_2}$$

Answer

$$\mathrm{T}=\mathrm{T}_1-\mathrm{T}_2$$

Oscillations

A particle connected to the end of a spring executes S.H.M. with period '$T_1'. While the corresponding period for another spring is '\mathrm{T}_2$'. If the period of oscillation with two springs in series is 'T', then

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