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MHT CET · Physics · Oscillations

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

  1. A \(\frac{\mathrm{T}_1+\mathrm{T}_2}{\mathrm{~T}_1 \mathrm{~T}_2}\)
  2. B \(\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1 \mathrm{~T}_2}\)
  3. C \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}\)
  4. D \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\mathrm{~T}_1+\mathrm{T}_2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}\)

Step-by-step Solution

Detailed explanation

\(
\mathrm{F}_1=\mathrm{K}_1 \mathrm{X} \text { and } \mathrm{F}_2=\mathrm{K}_2 \mathrm{X}
\)
When forces act simultaneously
\(\mathrm{F}=\left(\mathrm{K}_1+\mathrm{K}_2\right) \mathrm{x} \)
\( \mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_1}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_2}} \)
\( \mathrm{~T}^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{K}_1+\mathrm{K}_2} \)
\( \frac{1}{\mathrm{~T}^2}=\frac{\mathrm{K}_1+\mathrm{K}_2}{4 \pi^2 \mathrm{~m}}=\frac{\mathrm{K}_1}{4 \pi^2 \mathrm{~m}}+\frac{\mathrm{K}_2}{4 \pi^2 \mathrm{~m}}=\frac{1}{\mathrm{~T}_1^2}+\) \(\frac{1}{\mathrm{~T}_2^2}=\frac{\mathrm{T}_1^2+\mathrm{T}_2^2}{\mathrm{~T}_1^2 \mathrm{~T}_2^2} \)
\( \mathrm{~T}=\frac{\mathrm{T}_1 \mathrm{~T}_2}{\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}}\)
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