Question 333: Oscillations

Question

A body executes SHM under the action of force '$$\mathrm{F}_1$$' with time period '$$\mathrm{T}_1$$'. If the force is changed to '$$\mathrm{F_2}$$', it executes SHM with period '$$\mathrm{T_2}$$'. If both the forces '$$\mathrm{F_1}$$' and '$$\mathrm{F}_2$$' act simultaneously in the same direction on the body, its time period is

Accepted Answer

$$\begin{aligned}
	herefore \quad \mathrm{T}_1=\mathrm{K}_1 \mathrm{x} & =2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \quad \mathrm{F}_2=\mathrm{K}_2 \mathrm{x}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \quad \mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}_2ight) \mathrm{x} \
	herefore \quad \mathrm{T}_1^2 & =4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1} \quad \mathrm{~T}_2^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_2} \
\mathrm{~T}^2 & =4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1+\mathrm{K}_2} \
	herefore \quad \frac{1}{\mathrm{~T}^2} & =\frac{\mathrm{K}_1+\mathrm{K}_2}{4 \pi^2 \mathrm{~m}}=\frac{\mathrm{K}_1}{4 \pi^2 \mathrm{~m}}+\frac{\mathrm{K}_2}{4 \pi^2 \mathrm{~m}} \
& =\frac{1}{\mathrm{~T}_1^2}+\frac{1}{\mathrm{~T}_2^2} \
& =\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2} \
	herefore \quad \mathrm{T}^2 & =\frac{\mathrm{T}_1^2 \mathrm{~T}_2^2}{\mathrm{~T}_1^2+\mathrm{T}_2^2} \
	herefore \mathrm{T} & =\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}
\end{aligned}$$

Answer

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

Answer

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

Answer

$$\frac{\sqrt{T_1^2+\mathrm{T}_2^2}}{\mathrm{~T}_1 \mathrm{~T}_2}$$

Answer

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

Oscillations

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