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MHT CET · Physics · Oscillations

A particle connected to the end of a spring executes S.H.M. with period ' \(\mathrm{T}_1\) '. While the corresponding period for another spring is ' \(T_2\) '. If the period of oscillation with two springs in series is ' \(T\) ', then

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

Answer & Solution

Correct Answer

(A) \(\mathrm{T}=\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}\)

Step-by-step Solution

Detailed explanation

\(
\mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_1}} \quad \mathrm{~T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}_2}}
\)
When the two springs are connected in series, the effective spring constant is given by \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}\)
\(
\begin{aligned}
& \therefore \mathrm{T}^2=4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}}=4 \pi^2 \mathrm{~m}\left[\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}\right]=4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_1}+4 \pi^2 \frac{\mathrm{m}}{\mathrm{k}_2} \\
& =\mathrm{T}_1^2+\mathrm{T}_2^2 \\
& \therefore \mathrm{T}=\sqrt{\mathrm{T}_1^2+\mathrm{T}_2^2}
\end{aligned}
\)
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