MHT CET · Physics · Oscillations
A spring has a certain mass suspended from it and its period for vertical oscillations is ' \(T_1\) '. The spring is now cut in to two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now ' \(T_2\) '. The ratio \(T_1 / T_2\) is
- A \(2\)
- B \(\sqrt{2}\)
- C \(\frac{1}{\sqrt{2}}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_1=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} .\)
Also, spring constant \((\mathrm{k}) \propto \frac{1}{\text { Length }(l)}\)
When the spring is half in length, then \(\mathrm{k}\) becomes twice.
\(\begin{aligned}
& \therefore \quad \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \\
& \therefore \quad \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}
\end{aligned}\)
Also, spring constant \((\mathrm{k}) \propto \frac{1}{\text { Length }(l)}\)
When the spring is half in length, then \(\mathrm{k}\) becomes twice.
\(\begin{aligned}
& \therefore \quad \mathrm{T}_2=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}} \\
& \therefore \quad \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{1}{\frac{1}{\sqrt{2}}}=\sqrt{2}
\end{aligned}\)
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