MHT CET · Physics · Oscillations
The maximum velocity of a particle, executing S.H.M. with an amplitude 7 mm is \(4.4 \mathrm{~ms}^{-1}\). The period of oscillation is \(\left[\pi=\frac{22}{7}\right]\).
- A 100 s
- B 10 s
- C 0.1 s
- D \(\quad 0.01 \mathrm{~s}\)
Answer & Solution
Correct Answer
(D) \(\quad 0.01 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{ll}
& \mathrm{V}_{\max }=\mathrm{A} \omega \\
\therefore \quad & 4.4=7 \times 10^{-3}\left(\frac{2 \pi}{\mathrm{~T}}\right) \\
\therefore \quad & \mathrm{T}=\frac{7 \times 10^{-3} \times 2 \times 22}{4.4 \times 7}=0.01 \mathrm{~s}
\end{array}\)
\(\begin{aligned}
& \frac{mv^2}{r}= \frac{G M m}{r^2} \\
& v=\sqrt{\frac{G M}{r}} \\
& L=m v r \\
&=m \sqrt{\frac{G M}{r}} r \\
&=m(G M r)^{1 / 2}
\end{aligned}\)
& \mathrm{V}_{\max }=\mathrm{A} \omega \\
\therefore \quad & 4.4=7 \times 10^{-3}\left(\frac{2 \pi}{\mathrm{~T}}\right) \\
\therefore \quad & \mathrm{T}=\frac{7 \times 10^{-3} \times 2 \times 22}{4.4 \times 7}=0.01 \mathrm{~s}
\end{array}\)
\(\begin{aligned}
& \frac{mv^2}{r}= \frac{G M m}{r^2} \\
& v=\sqrt{\frac{G M}{r}} \\
& L=m v r \\
&=m \sqrt{\frac{G M}{r}} r \\
&=m(G M r)^{1 / 2}
\end{aligned}\)
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