MHT CET · Physics · Oscillations
A horizontal platform with a small object placed on it executes a linear S.H.M. in the vertical direction. The amplitude of oscillation is 40 cm . What should be the least period of these oscillations, so that the object is not detached from the platform? [Take \(g=10 \mathrm{~m} / \mathrm{s}^2\) ]
- A \(0.2 \pi \mathrm{~s}\)
- B \(0.3 \pi \mathrm{~s}\)
- C \(0.4 \pi \mathrm{~s}\)
- D \(0.5 \pi \mathrm{~s}\)
Answer & Solution
Correct Answer
(C) \(0.4 \pi \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
From the diagram,
\(\mathrm{mg} \div \mathrm{N}=\mathrm{ma}\)
For the object to just lose the Contact with the surface,
\(\begin{array}{ll}
& N=0 \\
\therefore \quad & m g=m a \\
\therefore \quad & g=a \\
\therefore \quad & g=A \omega^2
\end{array}\)
\(\begin{aligned} & \therefore \quad \omega=\sqrt{\frac{g}{A}}=\sqrt{\frac{10}{0.4}}=\sqrt{\frac{100}{4}}=\frac{10}{2}=5 \\ & \therefore \quad \frac{2 \pi}{T}=5 \\ & \therefore \quad T=\frac{2 \pi}{5}=0.4 \pi\end{aligned}\)
\(\mathrm{mg} \div \mathrm{N}=\mathrm{ma}\)
For the object to just lose the Contact with the surface,
\(\begin{array}{ll}
& N=0 \\
\therefore \quad & m g=m a \\
\therefore \quad & g=a \\
\therefore \quad & g=A \omega^2
\end{array}\)
\(\begin{aligned} & \therefore \quad \omega=\sqrt{\frac{g}{A}}=\sqrt{\frac{10}{0.4}}=\sqrt{\frac{100}{4}}=\frac{10}{2}=5 \\ & \therefore \quad \frac{2 \pi}{T}=5 \\ & \therefore \quad T=\frac{2 \pi}{5}=0.4 \pi\end{aligned}\)
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