Question 347: Oscillations

Question

A body performs S.H.M. under the action of force '$$\mathrm{F}_1$$' with period '$$\mathrm{T}_1$$' second. If the force is changed to '$$\mathrm{F}_2$$' it performs S.H.M. with period '$$\mathrm{T_2}$$' second. If both forces '$$\mathrm{F_1}$$' and '$$\mathrm{F_2}$$' act simultaneously in the same direction on the body, the period in second will be

Accepted Answer

$$\begin{aligned} & \mathrm{F}_1=\mathrm{K}_1 \mathrm{x} \quad \mathrm{F}_2=\mathrm{K}_2 \mathrm{x} \quad \mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}_2ight) \mathrm{x} \ & 	herefore T_1=2 \pi \sqrt{\frac{m}{K}} \quad T_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}}} \ & 	herefore \mathrm{T}_1^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1} \quad \mathrm{~T}_2^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_2} \ & T^2=4 \pi^2 \frac{m}{K_1+K_2} \ & 	herefore \frac{1}{\mathrm{~T}^2}=\frac{\mathrm{K}_1+\mathrm{K}_2}{4 \pi^2 \mathrm{~m}} \quad=\frac{\mathrm{K}_1}{4 \pi^2 \mathrm{~m}}+\frac{\mathrm{K}_2}{4 \pi^2 \mathrm{~m}} \ & =\frac{1}{\mathrm{~T}_1^2}+\frac{1}{\mathrm{~T}_2^2}=\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2} \ & 	herefore \mathrm{T}^2=\frac{\mathrm{T}_1^2 \mathrm{~T}_2^2}{\mathrm{~T}_1^2+\mathrm{T}_2^2} \ & 	herefore \mathrm{T}=\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}} \ & \end{aligned}$$

Answer

$$\frac{T_1+T_2}{T_1 T_2}$$

Answer

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

Answer

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

Answer

$$\frac{T_1 T_2}{T_1+T_2}$$

Oscillations

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