MHT CET · Physics · Oscillations
A body executes SHM under the action of force ' \(F_1\) ' 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
- A \(\frac{\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}}{\mathrm{~T}_1 \mathrm{~T}_2}\)
- B \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}}\)
- C \(\frac{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}{\mathrm{~T}_1 \mathrm{~T}_2}\)
- D \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{F}_1=\mathrm{K}_1 \mathrm{x} \quad \mathrm{F}_2=\mathrm{K}_2 \mathrm{X} \quad \mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}\right. \\ & \therefore \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_1}} \quad \mathrm{~T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_2}} \\ & \therefore \mathrm{T}_1^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1} \quad \mathrm{~T}_2^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_2} \\ & \mathrm{~T}^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1+\mathrm{K}_2} \\ & \therefore \frac{1}{\mathrm{~T}^2}=\frac{\mathrm{K}_1+\mathrm{K}_2}{4 \pi^2 \mathrm{~m}}=\frac{\mathrm{K}_1}{4 \pi^2 \mathrm{~m}}+\frac{\mathrm{K}_2}{4 \pi^2 \mathrm{~m}} \\ & =\frac{1}{\mathrm{~T}_1^2}+\frac{1}{\mathrm{~T}_2^2} \\ & =\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2} \\ & \therefore \mathrm{T}=\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}\end{aligned}\)
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