MHT CET · Physics · Oscillations
A particle of mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude \(0.3 \mathrm{~m}\) and period \(\frac{\pi}{5} \mathrm{~s}\). The maximum value of the force acting on the particle is
- A \(0.15 \mathrm{~N}\)
- B \(4 \mathrm{~N}\)
- C \(5 \mathrm{~N}\)
- D \(0.3 \mathrm{~N}\)
Answer & Solution
Correct Answer
(A) \(0.15 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \mathrm{m}=5 \mathrm{~g}=5 \times 10^{-3} \mathrm{~kg}, \mathrm{~A}=0.3 \mathrm{~m}, \mathrm{~T}=\frac{\pi}{5} \mathrm{~s} \\
& \omega=\frac{2 \pi}{\mathrm{T}}=10 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Maximum force, \(\mathrm{F}=\mathrm{m} \omega^2 \mathrm{~A}=5 \times 10^{-3} \times(10)^2 \times 0.3=0.15 \mathrm{~N}\)
\begin{aligned}
& \mathrm{m}=5 \mathrm{~g}=5 \times 10^{-3} \mathrm{~kg}, \mathrm{~A}=0.3 \mathrm{~m}, \mathrm{~T}=\frac{\pi}{5} \mathrm{~s} \\
& \omega=\frac{2 \pi}{\mathrm{T}}=10 \mathrm{rad} / \mathrm{s}
\end{aligned}
\)
Maximum force, \(\mathrm{F}=\mathrm{m} \omega^2 \mathrm{~A}=5 \times 10^{-3} \times(10)^2 \times 0.3=0.15 \mathrm{~N}\)
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