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A body performs S.H.M. under the action of force ' $F_1$ ' with period ' $T_1$ ' second. If the force is changed to ' $F_2$ ' it performs S.H.M with period ' $T_2$ ' second. If both forces ' $F_1$ ' and ' $F_2$ ' act simultaneously in the same direction on the body, the period in second will be
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Verified Answer
The correct answer is:
$\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}$
$$
\mathrm{F}_1=\mathrm{K}_1 \mathrm{X} \text { and } \mathrm{F}_2=\mathrm{K}_2 \mathrm{X}
$$
When forces act simultaneously
$$
\begin{aligned}
& \mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}_2\right) \mathrm{x} \\
& \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_1}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_2}} \\
& \mathrm{~T}^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1+\mathrm{K}_2} \\
& \frac{1}{\mathrm{~T}^2}=\frac{\mathrm{K}_1+\mathrm{K}_2}{4 \pi^2 \mathrm{~m}}=\frac{\mathrm{K}_1}{4 \pi^2 \mathrm{~m}}+\frac{\mathrm{K}_2}{4 \pi^2 \mathrm{~m}}=\frac{1}{\mathrm{~T}_1^2}+\frac{1}{\mathrm{~T}_2^2}=\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2} \\
& \mathrm{~T}=\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}
\end{aligned}
$$
\mathrm{F}_1=\mathrm{K}_1 \mathrm{X} \text { and } \mathrm{F}_2=\mathrm{K}_2 \mathrm{X}
$$
When forces act simultaneously
$$
\begin{aligned}
& \mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}_2\right) \mathrm{x} \\
& \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_1}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_2}} \\
& \mathrm{~T}^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1+\mathrm{K}_2} \\
& \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} \\
& \mathrm{~T}=\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}
\end{aligned}
$$
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