MHT CET · Physics · Oscillations
The displacement of a particle performing S.H.M is given by \(x=5 \sin (3 t+3)\), where \(x\) is in \(\mathrm{cm}\) and \(t\) is in second. The maximum acceleration of the particle will be
- A \(15 \mathrm{~cm} \mathrm{~s}^{-2}\)
- B \(30 \mathrm{~cm} \mathrm{~s}^{-2}\)
- C \(45 \mathrm{~cm} \mathrm{~s}^{-2}\)
- D \(90 \mathrm{~cm} \mathrm{~s}^{-2}\)
Answer & Solution
Correct Answer
(C) \(45 \mathrm{~cm} \mathrm{~s}^{-2}\)
Step-by-step Solution
Detailed explanation
\(x=5 \sin (3 t+3)\)
Standard equation of S.H.M. is \(x=A \sin (\omega t+\alpha)\)
\(\therefore \mathrm{A}=5 \mathrm{~cm}, \omega=3 \mathrm{rad} / \mathrm{s}\)
Maximum acceleration \(\mathrm{a}_{\mathrm{m}}=\mathrm{A} \omega^2=5 \times(3)^2\)
\(=5 \times 9=45 \mathrm{~cm} \mathrm{~s}^{-2}\)
Standard equation of S.H.M. is \(x=A \sin (\omega t+\alpha)\)
\(\therefore \mathrm{A}=5 \mathrm{~cm}, \omega=3 \mathrm{rad} / \mathrm{s}\)
Maximum acceleration \(\mathrm{a}_{\mathrm{m}}=\mathrm{A} \omega^2=5 \times(3)^2\)
\(=5 \times 9=45 \mathrm{~cm} \mathrm{~s}^{-2}\)
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