MHT CET · Physics · Oscillations
A child is sitting on a swing which performs S.H.M. It has minimum and maximum heights from ground \(0.75 \mathrm{~cm}\) and \(2 \mathrm{~m}\) respectively. Its maximum speed will be \(\left[\mathrm{g}=10 \frac{\mathrm{m}}{\mathrm{s}^2}\right]\)
- A \(\sqrt{1.25} \mathrm{~m} / \mathrm{s}\)
- B \(\sqrt{12.5} \mathrm{~m} / \mathrm{s}\)
- C \(5 \mathrm{~m} / \mathrm{s}\)
- D \(25 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(5 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Maximum potential energy is attained at the highest point which gets converted into kinetic energy at the lowest point.
\(
\begin{aligned}
& \mathrm{h}=2-0.75=1.25 \mathrm{~m} \\
& \frac{1}{2} \mathrm{mv}^2=\mathrm{mgh} \\
& \therefore \mathrm{v}^2=2 \mathrm{gh}=2 \times 10 \times 1.25=25 \\
& \therefore \mathrm{v}=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{h}=2-0.75=1.25 \mathrm{~m} \\
& \frac{1}{2} \mathrm{mv}^2=\mathrm{mgh} \\
& \therefore \mathrm{v}^2=2 \mathrm{gh}=2 \times 10 \times 1.25=25 \\
& \therefore \mathrm{v}=5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
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