MHT CET · Physics · Oscillations
Two identical springs of constant ' \(\mathrm{K}\) ' are connected in series and parallel in shown in figure. A mass ' \(\mathrm{M}\) ' is suspended from them. The ratio of their frequencies is series to parallel combination will be
- A 1:2
- B 1:4
- C 4:1
- D \(1: \sqrt{2}\)
Answer & Solution
Correct Answer
(A) 1:2
Step-by-step Solution
Detailed explanation
In series combination, the effective spring constant
\(
\mathrm{k}_1=\frac{\mathrm{k}}{2}
\)
In the parallel combination, the effective spring constant is
\(
\mathrm{k}_2=2 \mathrm{k}
\)
\(
\begin{aligned}
& \mathrm{f}_1=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{2 \mathrm{~m}}} \\
& \mathrm{f}_2=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}} \\
& \therefore \frac{\mathrm{f}_1}{\mathrm{f}_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\end{aligned}
\)
\(
\mathrm{k}_1=\frac{\mathrm{k}}{2}
\)
In the parallel combination, the effective spring constant is
\(
\mathrm{k}_2=2 \mathrm{k}
\)
\(
\begin{aligned}
& \mathrm{f}_1=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{2 \mathrm{~m}}} \\
& \mathrm{f}_2=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}} \\
& \therefore \frac{\mathrm{f}_1}{\mathrm{f}_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\end{aligned}
\)
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